url
stringlengths 6
1.61k
| fetch_time
int64 1,368,856,904B
1,726,893,854B
| content_mime_type
stringclasses 3
values | warc_filename
stringlengths 108
138
| warc_record_offset
int32 9.6k
1.74B
| warc_record_length
int32 664
793k
| text
stringlengths 45
1.04M
| token_count
int32 22
711k
| char_count
int32 45
1.04M
| metadata
stringlengths 439
443
| score
float64 2.52
5.09
| int_score
int64 3
5
| crawl
stringclasses 93
values | snapshot_type
stringclasses 2
values | language
stringclasses 1
value | language_score
float64 0.06
1
|
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
https://quant.stackexchange.com/questions/22564/americanoptionimpliedvolatility-root-not-bracketed-issue-in-quantlib-r
| 1,620,467,755,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243988858.72/warc/CC-MAIN-20210508091446-20210508121446-00015.warc.gz
| 491,593,713
| 38,440
|
# AmericanOptionImpliedVolatility - root not bracketed issue in QuantLib/R
I'm trying to compute an implied volatility -- I am trying to match real data I see in Yahoo finance which shows an IV of about 27%. My call in 'R' for the same params returns a root not bracketed error -- can anyone help pls?
Sample:
> AmericanOptionImpliedVolatility(type="put", value=2.7,
underlying=55.0, strike=60, dividendYield=0.02, riskFreeRate=0.03,
maturity=0.02, volatility=0.2)
RESULT:
Error in americanOptionImpliedVolatilityEngine(type, value, underlying,
: root not bracketed: f[1e-07,4] -> [2.300000e+00,1.256782e+01]
It is a simple root finder, and if you give it impossible starting values... well then it fails. Here, you can play with the values and it seems bounded at USD 5 whereas you start from USD 2.7:
R> AmericanOption(type="put", underlying = 55, strike = 60,
+ dividendYield = 0.02, riskFreeRate = 0.03,
+ maturity = 0.02, volatility = 0.2)
Concise summary of valuation for AmericanOption
value delta gamma vega theta rho divRho
5 NA NA NA NA NA NA
R>
Maybe you had strike and underlying mixed up?
R> AmericanOptionImpliedVolatility(type="put", value=2.7,
+ underlying=60.0, strike=55.0, dividendYield=0.02,
+ riskFreeRate=0.03, maturity=0.02, volatility=0.2)
[1] 1.48203
attr(,"class")
[1] "AmericanOptionImpliedVolatility" "ImpliedVolatility"
R>
• Yes. There's no way that a put option with underlying at 55 and strike at 60 has a value of 2.7. It has to be worth at least 5, because that's what you would gain by exercising now. – Luigi Ballabio Jan 2 '16 at 22:04
• That helps -- thanks. But let's try to look at a specific example as I'm still stuck trying to match what Yahoo shows: code From Yahoo -- today -- an MSFT 40 PUT, for July 15, 2016 has the following (bid=0.38, ask=0.40) - IV = 31.86% code > AmericanOptionImpliedVolatility(type="put", value=0.39, underlying=55.48, strike=40, dividendYield=0.00, riskFreeRate=0.01, maturity=0.53, volatility=0.2) [1] 0.3224758 attr(,"class") ' So this matches. – nxstock-trader Jan 2 '16 at 22:25
| 661
| 2,154
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2021-21
|
longest
|
en
| 0.832172
|
https://www.eduzip.com/ask/question/displaystyle-int-pi4pi4-ln-sqrt1sin2xdx-dx-has-the-value-equal-to-580132
| 1,623,786,265,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487621519.32/warc/CC-MAIN-20210615180356-20210615210356-00019.warc.gz
| 683,702,610
| 8,493
|
Mathematics
# $\displaystyle \int_{\pi/4}^{\pi/4} ln \sqrt{1+\sin2x}dx$ dx has the value equal to:
$\dfrac{\pi}{4}\ln2$
Its FREE, you're just one step away
Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
#### Realted Questions
Q1 Single Correct Medium
If $f(x)=\displaystyle \frac{e^{x}}{1+e^{x}}$ ,$I_{1}=\displaystyle \int_{f(-a)}^{f(a)}xg\{x(1-x)\}dx$ and $I_{2}=\displaystyle \int_{f(-a)}^{f(a)}g\{x(1-x)\}dx$, then the value $\displaystyle \frac{I_{2}}{I_{1}}$ is
• A. -3
• B. -1
• C. 1
• D. 2
1 Verified Answer | Published on 17th 09, 2020
Q2 Subjective Hard
Integrate
$\int {\frac{{2\sin 2x - \cos x}}{{6 - {{\cos }^2}x - 4{\mathop{\rm sinx}\nolimits} }}dx}$
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
$\int \cos(\ln x)dx=$
• A. $\dfrac{x}{2}[\cos \ln x-\sin \ln x]$
• B. $[x \cos \ln x + \sin \ln x]$
• C. None of these
• D. $\dfrac{x}{2}(\cos \ln x+\sin \ln x)$
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Hard
If $\displaystyle \int_{1}^{x}\dfrac{dt}{|t|\sqrt{t^{2}-1}}=\dfrac{\pi}{6}$, then $x$ can be equal to :
• A. $\sqrt{3}$
• B. $2$
• C. $\dfrac{4}{\sqrt{3}}$
• D. $\dfrac{2}{\sqrt{3}}$
$\int(2x^2-3 \, sin x+5 \sqrt{x})dx$
| 545
| 1,264
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.296875
| 3
|
CC-MAIN-2021-25
|
latest
|
en
| 0.500294
|
https://brilliant.org/problems/can-you-handle-this-one/
| 1,529,410,921,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267862929.10/warc/CC-MAIN-20180619115101-20180619135101-00007.warc.gz
| 581,981,349
| 11,226
|
# Can You Handle This One
Algebra Level 4
$\large{f(x,y,z)= 2x^{2}+2y^{2}-2z^{2}+\frac{7}{xy}+\frac{1}{z}}$
There are three pairwise distinct numbers $$a,b,c$$ that satisfies the above equation in a way such that $$f(a,b,c)= f(b,c,a)=f(c,a,b)$$.
What is $$a+b+c$$?
###### Problem Source: cms.math/competitions 2009.
×
Problem Loading...
Note Loading...
Set Loading...
| 124
| 375
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.9375
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.659793
|
https://www.distancesto.com/road-map/in/kadayanallur-to-kadambur/history/269723.html
| 1,643,185,832,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320304928.27/warc/CC-MAIN-20220126071320-20220126101320-00229.warc.gz
| 747,284,052
| 10,750
|
The road map above shows you the route to take to your destination. You can toggle between map views using the buttons above.
### Map Summary
Now that you've seen the map of your trip. You may want to know the distances from Kadayanallur to Kadambur? If so, see the distance from Kadayanallur to Kadambur.
Since this map only shows you the route of your trip and doesn't actually tell you how to get to your destination, you may want to see driving directions from Kadayanallur to Kadambur.
After figuring out how to get there, you'll also want to know the travel time since we all want to get to where we need to go quickly. Which is why you'll want to get a better idea of how long is takes by looking at the travel time from Kadayanallur to Kadambur.
Next you'll want to calculate the cost of the trip. How? By figuring out the fuel consumption of your vehicle and entering in the cost of the fuel. Calculate the fuel cost from Kadayanallur to Kadambur.
After figuring out the cost of your trip, you can also figure out the flight distance since some trips might be worth flying. To figure out whether they are worth it, see the flight distance from Kadayanallur to Kadambur.
Whether the distance is short or long, we can always figure out how long it would take you to fly there anyway by viewing the flight time from Kadayanallur to Kadambur.
| 310
| 1,354
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.546875
| 3
|
CC-MAIN-2022-05
|
latest
|
en
| 0.960602
|
https://lists.ufl.edu/cgi-bin/wa?A3=ind1102&L=SOCNET&E=8bit&P=725923&B=--&T=text%2Fplain;%20charset=ISO-8859-1&header=1
| 1,619,138,983,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618039563095.86/warc/CC-MAIN-20210422221531-20210423011531-00203.warc.gz
| 458,162,741
| 2,253
|
```***** To join INSNA, visit http://www.insna.org *****
Tad, do you imply that a solution is known for k = 1? Isn't this NP? --Moses
2011/2/23 <[log in to unmask]>:
> ***** To join INSNA, visit http://www.insna.org *****
>
> Dear Socnet Members
>
> Let me address a question to those of you who have been watching
> developments in GRAPH THEORY.
>
> Is the following statement true?
>
> 'No nontrivial complete set collection of invariants has yet been found
> for directed graphs
>
> Definitions
>
> A (finite) DIRECTED GRAPH is a mathematical object of the form G=(X,R),
> where X is a finite set (of points) and R is a binary relation in X.
>
> Two digraphs G=(X,R) and G =(X ,R ) are ISOMORPHIC if there exists a 1-1
> mapping F of X onto X such that xRy if and only if F(x)R F(y), for any
> x,y in X.
>
> A STRUCTURAL PARAMETER of a digraph is any mapping P which assigns a
> numerical value to any digraph in such a way that P(G)=P(G ) if G and G
> are isomorphic.
>
> A COMPLETE SET OF INVARIANTS is any finite collection P1, ,Pk of
> structural parameters such that, for any two digraphs G and G , the
> equations P1(G)=P1(G ), ,Pk(G)=Pk(G ) imply that G and G are isomorphic.
>
> For any n, let gn denote the number of equivalence classes of the
> isomorphism relation in the set of digraphs with X={1, ,n}. The numbers
> 1, ,gn can be used as code symbols of equivalence classes. A 'trivial'
> set of invariants is obtained by assigning to any G=(X,R) the number n of
> points (n=|X| ) and an appropriate symbol from the range from 1 to gn.
> Hence, by saying that a set of invariants is NONTRIVIAL I mean that the
> values of P1, ,Pk for two n-point digraphs G and G can be computed
> without prior knowledge of the list of isomorphism classes.
>
> Tad Sozanski
> www.cyf-kr.edu.pl/~ussozans/
>
> _____________________________________________________________________
> SOCNET is a service of INSNA, the professional association for social
> network researchers (http://www.insna.org). To unsubscribe, send
> an email message to [log in to unmask] containing the line
> UNSUBSCRIBE SOCNET in the body of the message.
>
_____________________________________________________________________
SOCNET is a service of INSNA, the professional association for social
network researchers (http://www.insna.org). To unsubscribe, send
an email message to [log in to unmask] containing the line
UNSUBSCRIBE SOCNET in the body of the message.
```
| 698
| 2,468
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.96875
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.821069
|
https://www.circuitstoday.com/voltage-comparator
| 1,718,819,959,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861828.24/warc/CC-MAIN-20240619154358-20240619184358-00533.warc.gz
| 627,907,515
| 18,430
|
## Voltage comparator circuit.
Voltage comparator is a circuit which compares two voltages and switches the output to either high or low state depending upon which voltage is higher. A voltage comparator based on opamp is shown here. Fig1 shows a voltage comparator in inverting mode and Fig shows a voltage comparator in non inverting mode.
### Non inverting comparator.
In non inverting comparator the reference voltage is applied to the inverting input and the voltage to be compared is applied to the non inverting input. Whenever the voltage to be compared (Vin) goes above the reference voltage , the output of the opamp swings to positive saturation (V+) and vice versa. Actually what happens is that, the difference between Vin and Vref, (Vin – Vref) will be a positive value and is amplified to infinity by the opamp. Since there is no feedback resistor Rf, the opamp is in open loop mode and so the voltage gain (Av) will be close to infinity. So the output voltage swings to the maximum possible value ie; V+. Remember the equation Av = 1 + (Rf/R1). When the Vin goes below Vref, the reverse occurs.
### Inverting comparator.
In the case of an inverting comparator, the reference voltage is applied to the non inverting input and voltage to be compared is applied to the inverting input. Whenever the input voltage (Vin) goes above the Vref, the output of the opamp swings to negative saturation. Here the difference between two voltages (Vin-Vref) is inverted and amplified to infinity by the opamp. Remember the equation Av = -Rf/R1. The equation for voltage gain in the inverting mode is Av = -Rf/R1.Since there is no feedback resistor, the gain will be close to infinity and the output voltage will be as negative as possible ie; V-.
### Practical voltage comparator circuit.
A practical non inverting comparator based on uA741 opamp is shown below. Here the reference voltage is set using the voltage divider network comprising of R1 and R2. The equation is Vref = (V+/ (R1 + R2)) x R2. Substituting the values given in the circuit diagram into this equation gives Vref = 6V. Whenever Vin goes above 6V, the output swings to ~+12V DC and vice versa. The circuit is powered from a +/- 12V DC dual supply.
Few other opamp related circuits that you may be interested in.
Integrator using opamp : For an integrating circuit, the output signal will be the integral of the input signal. For example, a sine wave on integration gives cosine wave, square wave on integration gives triangle wave etc.
Inverting amplifier : In an inverting amplifier, the output signal will be an inverted version of the input signal and is amplified by a certain factor.
Instrumentation amplifier : This is a type of differential amplifier with additional buffer stages at the input. This results in high input impedance and easy matching. The instrumentation amplifier has better stability, high CMRR, low offset voltage and high gain.
1. Gary Metz
Can some one please tell me how to use comparators to control the out put voltage of a generator which uses a PWM to power the field windings. The needed out put is 220VAC and the voltage to the fields is between 40 and 90 depending on the load. Now it is controlled by turning a pot on the PWM but needs to be automatic. Also the out put voltage needs to be changed at times to suit other needs, as low as 56v.
2. vitthal Dhone
explained in simple way , previously, i didn’t i understand but now i got .
3. FAWAZ,ABDULLAHIL
it’s something good for us.we can learn a lot of things about comparators.
4. its awsome but practical comparator circuit is not showing output in multisim 🙁
5. it was useful, thanks
i had completely wrong notes.now i feel better 🙂
| 832
| 3,716
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.390625
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.858392
|
https://www.proprofs.com/quiz-school/story.php?title=NjEyNDAz
| 1,653,418,128,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662573189.78/warc/CC-MAIN-20220524173011-20220524203011-00687.warc.gz
| 1,095,094,080
| 51,935
|
# Ray Optics And Optical Instruments
15 Questions | Total Attempts: 1610
Settings
Time: 30 Minute
• 1.
The absolute refractive indices of water and glass are 1.3 and 1.5 respectively. The refractive index of water w.r.t. glass is
• A.
• B.
• C.
• D.
None of these
• 2.
The focal length of the lens is 50 cm; then its power is
• A.
+ 1 D
• B.
+ 2 D
• C.
– 2 D
• D.
– 1 D
• 3.
A convex mirror has a focal length f. A real object, placed at a distance f in front of it from the pole, produces an image at
• A.
F
• B.
• C.
2f
• D.
Infinity
• 4.
A convex lens of focal length x and a concave lens of focal length y are placed in contact. The focal length of the combination is
• A.
X + y
• B.
X - y
• C.
• D.
• 5.
A graph is plotted between the angle of incidence (i) and angle of reflection (r). The correct variation is shown by
• A.
• B.
• C.
• D.
• 6.
If a water drop is kept between two glass plates, then its surface is correctly shown by
• A.
• B.
• C.
• D.
• 7.
Figure shows three cases of a ray passing through a prism of refracting edge A. The case corresponding to minimum deviation is
• A.
• B.
• C.
• D.
None of these
• 8.
A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach the face BC if:
• A.
Sin C = 8/9
• B.
Sin C = 9/8
• C.
Sin C = 2/3
• D.
Sin C = 3/2
• 9.
In an experiment to find focal length of a concave mirror, a graph is drawn between the magnitude of u and v. The graph looks like:
• A.
• B.
• C.
• D.
• 10.
What should be the refractive index of a completely transparent medium for it to be visible in vacuum?
• A.
1
• B.
Less than 1
• C.
Greater than 1
• D.
None of the above
• 11.
Rainbow is formed due to combination of:
• A.
Refraction and absorption
• B.
Dispersion and foccusing
• C.
Refraction and scattering
• D.
Dispersion and total internal reflection
• 12.
The plane face of a plano-convex lens of focal length 20 cm is silvered as shown in Figure. What kind of a mirror will it behave?
• A.
Convex, f = 20 cm
• B.
Convex, f = 10 cm
• C.
Concave, f = 20 cm
• D.
Concave, f = 10 cm
• 13.
A telescope has an objective lens of 10 cm diameter and is situated at a distance of 1 km from two objects. The minimum distance between theses two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Ǻ is of order of:
• A.
5 mm
• B.
5 cm
• C.
2.5 m
• D.
5 m
• 14.
Focal length of a glass lens in air is 2 cm. Its focal length when immersed in water would be
• A.
4cm
• B.
6cm
• C.
8 cm
• D.
12 cm
• 15.
If a beam of light passes from air to glass, the speed of light
• A.
Decreases
• B.
Increases
• C.
Remains unchanged
• D.
It may increases or decrease depending upon the colour
Related Topics
| 884
| 2,843
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.841752
|
http://tutorial.wmlcloud.com/windows_phone/Developing-for-Windows-Phone-and-Xbox-Live---Projection-Matrix.aspx
| 1,618,607,032,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038089289.45/warc/CC-MAIN-20210416191341-20210416221341-00064.warc.gz
| 104,093,211
| 9,972
|
Windows Phone
# Developing for Windows Phone and Xbox Live : Projection Matrix
5/30/2011 11:44:12 AM
Compared to the view matrix, the projection matrix is much more difficult to explain! When thinking of the view matrix as the camera, you can think of the projection matrix as the lens. There are many more lens types than cameras!
#### Perspective
The perspective matrix lets you define the viewing frustum. To visualize what this is, imagine a pyramid. The tip of the pyramid is the location of the camera (or your eye). Anything that exists inside that pyramid can be seen; anything that exists outside of the pyramid cannot be seen. The various methods to create perspective projection matrices are used to define the size and shape of this pyramid. See Figure 1.
##### Figure 1. The viewing frustum
First, let’s look at the matrix used in the example. It was created using the CreatePerspectiveFieldOfView method, which has a few parameters. The first is the field of view parameter, the second is the aspect ratio, and the last two are the near and far planes, but what do these actually mean? The field of view is the angle that the top of the pyramid forms and is specified in radians. In the example, we used an angle of 45 degrees, which is somewhat common. If you used 90 degrees (MathHelper.PiOver2) instead, you’d have a much larger field of view, and thus, a larger viewing area. This would cause your rendered objects to appear smaller because the same screen area would have to render a larger world area (see Figure 2 for more information on the math of the field of view or fov).
##### Figure 2. The math behind field of view
The second parameter is the aspect ratio, which is simply the width of your scene divided by the height of your scene. Notice here that you use the Viewport property of the GraphicsDevice and you use the AspectRatio property from that. This is the most common aspect ratio you use, but if you render to something other than the back buffer (say a render target), you might want to render at a different aspect ratio. The aspect ratio here is just like the aspect ratio you see on televisions. Standard-definition televisions have a 1.33 (4/3) aspect ratio, whereas most high-definition televisions have a 1.77 (16/9) aspect ratio. Most of the old resolutions you’ve seen in computers were 1.33 (4/3) aspect ratios (640×480, 800×600, 1024×768, and so on).
Aspect Ratio, Integers, and Floats
Aspect ratio is defined as a float value. If you are calculating your aspect ratio using the width divided by the height, and your width and height parameters are both integers, so be sure to cast at least one of them to float before the division. If you do not, the compiler does the division as integers and casts the final value to float, which isn’t what you want. For example, if you try to calculate the aspect ratio of a standard definition television set, you use the width of 4 divided by the height of 3, and in integer math, 4/3 = 1. It would then cast the 1 to the float of 1.0f, which is incorrect. If you cast one of the values to float before the division, though, you get 4.0f/3 = 1.3333f, which is the value you want.
The last two parameters are the near and the far planes of the frustum. Anything closer than the near plane or anything farther than the far plane is not visible. You can think of the far plane as the base of the pyramid, whereas the near plane is where you would “cut off” the top of the pyramid.
See Figure 2 for more information on the math for a field of view projection matrix.
Using the field of view is one way to create a perspective projection matrix, but not the only way. There are two other helper methods you can use, namely CreatePerspective and CreatePerspectiveOffCenter. Each of these creates your pyramid in slightly different ways, but before we get to that, let’s modify the example to draw a lot of boxes so it’s easier to see how the changes behave. Replace your draw code with the following (this might seem way more complicated than it is) to see a series of boxes like you see in Figure 3:
`protected override void Draw(GameTime gameTime){ const int numberBoxes = 3; const int radiusMultiple = numberBoxes + 1; GraphicsDevice.Clear(Color.CornflowerBlue); float radius = model.Meshes[0].BoundingSphere.Radius;Matrix view = Matrix.CreateLookAt( new Vector3(0, radius * radiusMultiple, radius * (radiusMultiple * radiusMultiple)), new Vector3((numberBoxes / 2) * (radius * radiusMultiple) - 1, (numberBoxes / 2) * (radius * radiusMultiple) - 1, 0), Vector3.Up);Matrix proj = Matrix.CreatePerspectiveFieldOfView(MathHelper.PiOver4, GraphicsDevice.Viewport.AspectRatio, 1.0f, 100.0f); for (int x = 0; x < numberBoxes; x++) { for (int y = 0; y < numberBoxes; y++) { Vector3 pos = new Vector3((y * (radius * radiusMultiple)) - 1, (x * (radius * radiusMultiple)) - 1, -(y + x)); model.Draw(Matrix.CreateTranslation(pos), view, proj); } } base.Draw(gameTime);} `
##### Figure 3. The view of a bunch of boxes
Although this seems more complicated than it is, all you’re doing here is drawing nine different boxes. They’re rendered in a square pattern with each one rendered at a different spot in the world and at a variety of different depths. You should notice how ones that are farther away are slightly smaller.
The CreatePerspective method takes only four parameters to describe what your pyramid should look like. The first one is the width of the top of the pyramid (where the near plane cuts it off), and the second is the height of the same. The third parameter is the distance to the near plane, and the fourth is the distance to the far plane (or the base of the pyramid). Modify your matrix creation as follows:
`Matrix proj = Matrix.CreatePerspective(1.0f, 0.5f, 1.0f, 100.0f);`
If you run the program now, notice that the picture has changed somewhat dramatically. All nine boxes don’t even appear fully onscreen anymore! This is because you changed the shape of your viewing frustum (or the pyramid), and portions of those boxes are now outside of it. See Figure 4.
##### Figure 4. Modifying the view frustum directly
The CreatePerspective method assumes that you are looking at the center point of the near plane (formed by the width and height), but that isn’t a requirement either. You can use the CreatePerspectiveOffCenter method, which is similar to the nonoff center method. Rather than a width and height being passed in, you instead must pass in the left, right, top, and bottom positions. For example, if you use the following instead of the CreatePerspective method you used earlier, you would get the same output:
`proj = Matrix.CreatePerspectiveOffCenter(-0.5f, 0.5f, -0.25f, 0.25f, 1.0f, 100.0f); `
This is because you have the same width and height, and they are centered. Using this method enables you even more control over the location and size of the viewing frustum.
There is also another common type of projection matrix, the orthographic projection.
#### Orthographic
Much like a perspective matrix, an orthographic matrix builds a viewing frustum, but instead of a pyramid shaped structure, it is more rectangular. The viewing frustum does not get larger between the near plane and the far plane, and no perspective foreshortening occurs. All objects of the same size are rendered the same size, regardless of how far they are from the camera.
There are two helper methods to create these types of projection matrix: CreateOrthographic and CreateOrthographicOffCenter. Much like the perspective counterparts, these describe the orthographic volume, with the former being centered and the latter capable of being off center. If you replaced your project matrix with the following, you would see all nine boxes, but they’d all appear on the same plane:
`proj = Matrix.CreateOrthographic(15.0f, 15.5f, 1.0f, 100.0f);`
With the basics of projection matrices and view matrices out of the way, now you can actually create some camera types!
Other -----------------
Top 10
| 1,896
| 8,166
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.15625
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.921599
|
https://calculator.name/weight/kip/milligram
| 1,718,980,791,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198862125.45/warc/CC-MAIN-20240621125006-20240621155006-00856.warc.gz
| 126,834,508
| 22,646
|
Amount
From
To
# How to convert kip to milligrams?
Kip and Milligrams are measuring units used to measure weights.
## To convert from kip to milligrams, multiply the number of kip by 453592370 and the result is in milligrams
### How to use kip to milligrams conversion calculator
1. Enter number of kip in the given input box.
2. Click on the "Calculate" button to convert number of kip to milligrams.
3. Change "From" or "To" units value for different weight units conversion tool
Kip Conversion of Kip to Milligrams
0.99 kip 0.99 kip = 449056446.3 mg
0.9 kip 0.9 kip = 408233133 mg
1 kip 1 kip = 453592370 mg
2 kip 2 kip = 907184740 mg
3 kip 3 kip = 1360777110 mg
| 212
| 670
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2024-26
|
latest
|
en
| 0.65973
|
https://bookofthrees.com/category/features/featured-articles/?page_number_0=2
| 1,537,325,148,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-39/segments/1537267155817.29/warc/CC-MAIN-20180919024323-20180919044323-00059.warc.gz
| 478,621,972
| 26,446
|
# Category Archives: Featured Articles
Featured articles are long and deserve more attention. They have been saved in PDF format and html formats so you may print them out. Please enjoy the rich, thoughtful and provoking articles.
# List-Construction as a Task and Resource
By Gail Jefferson
This report is a preliminary examination of lists occurring in natural conversation. It focuses upon the work which list-construction, as a task, allots to speakers, and some uses to which list-construction, as a resource, can be put by speakers.
The presence of three-part lists are first noted. Speakers and hearers orient to their three-part nature. The completed list can then constitute a turn at talk and the hearer can monitor the third component as a sign of turn completion. Lists can thereby’ be a conversational sequential resource.
By virtue of the three-part structure of some lists, members can orient to such matters as a “weak,” “absent,” or “missing” third part. Third items can be used to accomplish particular interactional work, such as topic-shifting and offense avoidance.
Further, a list can be constructed by more than one speaker. This feature may be used for a range of activities, including the achievement of interactional accord in situations of impending discord.
Read report here – List construction
# On the Nature of the Universe
### On the Nature of the Universe
By Scott Anfield
For many years, people have sought to discover a unified field theory which explains all the phenomena in the Universe, from the smallest particles to the galaxies and even the structure of the universe itself. The theory which I propose encompasses all of these.
This theory is concerned with the fact that all things come in threes, of which there are numerous examples (see Appendix). The whole of anything is comprised of 3 parts. All 3 of these parts have something in common, which gives the appearance of a whole. The 3 parts are divided into a ratio of 1:2, of which two are the same as each other in some respect and one is different. Herein lays the stability of this theory, as it must always be two of the same and one different because when reversed, the single part would have nothing to be the same as. However, all 3 parts are interlinking with each other. Each part has something in common with every other part and they both have a difference with the other part. For example, parts one, two and three can be shown as A, B and C respectively.
• A and B have a common link with each other, but both different from C
• A and C have a common link with each other, but both different from B
• B and C have a common link with each other, but both different from A.
I have named this tri-interconnectivity
# Three as a magic number in Latin Literature
Read Three as a magic number in Latin Literature (pdf file)
Or browse the Full text of “Three as a Magic Number in Latin Literature” by JStor.
Vol. xlvii] Three as a Magic Number in Latin Literature
By Professor EUGENE TAVENNER
MIDDLE TENNESSEE NORMAL SCHOOL
To one who is investigating the subject of magic among the Romans few references to magical details seem so per- sistently repeated as those concerning the number three. It is my desire, therefore, to present in this paper all the pas- sages referring to three as a magic number that I have been able to find in the literature of Rome up to the beginning of the fourth century of our era, with some additional illustrative material drawn from a later date. 1
# Living in Threes: Triangles and Triglyphs in Ancient Cultures
By Claire Holt
Few periods of ancient history sum up mathematical precision in quite as dramatic a fashion as Ancient Egypt. Against a rugged landscape of rocky mountains, rolling sand dunes, and the wide emptiness of an endless blue sky, the architects in the Land of the Pharaohs embraced geometric design with a passion unrivaled by any other civilization. The Pyramids at Giza remain one of the great architectural wonders of the world, and the giant sculpture of the Sphinx is an unrivaled example of the Egyptian ability to represent the natural form within a geometric methodology. Equally interested in the mystical power of numerology were the Ancient Greeks. With their elegant marble temples and fertile landscapes, the Greeks built a civilization of which the number three was an object of passion. Its legacy has continued to live on as a core element of more modern cultural codes and religions, suggesting that three may be more important to the way that we currently view the world than we necessarily realize.
# Ternarity in Thinking, Culture, and Art: System- informational Roots of Unconsciousness
Петров, В. М.
2008, vol. 5, No. 4, pp. 3–18
The results obtained in the framework of six informational models of perception prove the preference of encoding of input information by three-graded features, the preference of using of three-parameter mechanisms and reliable fixation of periodic events which are repeated three times. In the latter case the excess of chance threshold and fixation of regularity by means of positive emotions play an important role. Deduced ternarity dominates in the unconscious, it is revealed primarily in phenomena of culture (colour triads in national schools of painting, three stage structure of literature texts, religious, philosophical, language and other systems) as well as in three-factorial structure of semantic space and three-dimensionality of the perceived world.
Full Text here (In Russian)
# “Intensify this Awe”:Threeness, the Triad, and Christopher Alexander
[Original longer title: Identifying the Conceptual and Practical Power of Christopher Alexander’s Theory and Practice of Wholeness: Clues as Provided by British Philosopher J. G. Bennett’s Systematics of the Triad]
David Seamon
Department of Architecture
Kansas State University
Manhattan, KS 66506
www.arch.ksu.edu/seamon
Excerpt: |
Bennett provides a simple example of the triad of interaction in everyday experience: “I am sitting in my study on a cold winter evening and do not notice that the fire has burnt low until my body experiences a sensation of cold. My attention being thus drawn to the fire, I get up, take a poker and poke the fire. When I see that it is burning up, I return to my chair and continue reading.
“The whole event is a cycle of interactions, beginning and ending with the bodily sensations of cold and heat. It can be broken down into a series of triads, starting with my reaction to the sensation of cold. Here the physical sensation links the fall of temperature with my getting up and taking the poker. The environment is active and my body is passive; sensation is the reconciling impulse. When I get up and poke the fire, my body is active, the fire is passive, and the poker transmits the reconciling impulse. When I begin to feel warm again, the fire is active, my body is passive and the radiation of the fire and the wamr air of the room transmit the reconciling impulse.
“The roles of the different objects—air, body, poker, fire—change from one triad to the other. There is neither expansion nor concentration but a change in the distribution of energy. The event can be thus analyzed in greater or less detail, but it will always prove to consist of a nexus of triads in which one entity is acting on another through the medium of a third.
# Research into the “Threes” Phenomena – Herb O. Buckland
My views of the “Threes” phenomena have changed over the years. Whereas many decades ago I began collecting various patterns-of-three examples as if they constituted some fundamental universal law, I have come to realize that many of the so-called universals exist with what may be termed an auxiliary pattern. Namely, a 3-to-1 ratio that I alternatively describe as a 3-into-1, 3-from-1, or 3-as-1, though these proportions can be turned around to read 1-into-3, 1-from-3, or 1-as-3, whether one uses numbers, letters, symbols, sounds, etc., mix or match as you will. . .
The above paragraph is only an excerpt. Herb has an amazing site. I encourage you to read more at http://threesology.org. He also has laminated poster for sale! You may contact him directly at herbobuckland@hotmail.com.
# Triplicity – Simon Kelsey
What is Triplicity?
According to the Merriam-Webster dictionary Triplicity has two meanings or definitions:
1. Triplicity is one of the groups of three signs (each distant 120 degrees from the other two) into which the signs of the zodiac are divided.
2. Triplicity is the quality or state of being triple or threefold.
3. To these I propose adding a third meaning or definition: Triplicity is the phenomenon of Threeness in life.
Why a third one? Well let’s step back a bit and begin by defining the term “threes”. There are three primary colors red, blue and green, and three states of matter gas, liquid and solid. Space has three dimensions and we divide time into the past, present and future. In other words “threes” are groups of three things that are distinct from and yet related to each other.
What is really interesting is that there are numerous “threes” in every sphere of life. Threes are an observable phenomenon. I considered various names for this phenomenon and eventually settled on Triplicity, and so the third meaning or definition.
The Wikipedia entry on ternary numeral systems notes:
A base-three system is used in Islam to keep track of counting Tasbih to 99 or to 100 on a single hand for counting prayers (as alternative for the Misbaha). The mnemonic benefit is that counting within this system then reduces distraction since the counter needs only to divide Tasbihs into groups of three.
use of ternary numbers conveniently to convey self-similar structures like a Sierpinski Triangle or a Cantor set. The ternary representation is useful for defining the Cantor Set and related point sets, because of the way the Cantor set is constructed.
ternary as being the integer base with the highest radix economy, followed closely by binary and quaternary. It has been used for some computing systems because of this efficiency. Rarely mentioned is the existence of ternary computers (notably defining a tryte to be 6 trits, analogous to the binary byte).
use in the representation of 3 option trees, such as phone menu systems, which allow a simple path to any branch.
Of further relevance to the pattern of argument here is the role of ternary valued logic. Such a three-valued or trivalent logic is one in which there are three truth values indicating true, false and some third value. This is contrasted with the more common bivalent logics (mentioned above) which provide only for true and false. or guilty and not-guilty. An exception occurs in the Scottish legal system providing additionally for not-proven (a distinction which would seem to be of considerable current significance with respect to many detained in Guantanamo Bay).
Conceptual form and basic ideas were initially created by Jan Lukasiewicz, C. I. Lewis and Sulski. These were then re-formulated by Grigore Moisil in an axiomatic algebraic form, and also extended to n-valued logics. In the argument here, the question is whether the pattern in the diagram above holds a meaningful relationship with a range of multi-valued logic systems.
| 2,465
| 11,327
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.640625
| 3
|
CC-MAIN-2018-39
|
latest
|
en
| 0.952589
|
https://kr.mathworks.com/matlabcentral/answers/478709-can-you-help-me-please
| 1,601,276,757,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600401585213.82/warc/CC-MAIN-20200928041630-20200928071630-00710.warc.gz
| 461,965,760
| 23,904
|
# Can you help me please !
조회 수: 1(최근 30일)
I run code Matlab:
function yp=khidonghoc(t,y)
m = 10;
cd = 0.2;
g = 9.81;
w = m*g;
tspan=[0 5];
y0=[0;100;0;10];
yp = zeros(4,1);
yp(1)= y(2);
yp(2) = ((-cd/m)*y(2)*(y(2)^2+y(4)^2)^(0.5));
yp(3) = y(4);
yp(4) = ((-w/m)-(cd/m)*y(4)*(y(2)^2+y(4)^2)^(0.5));
[t,y]=ode45('khidonghoc',tspan,y0);
plot(t,y(:,1),y(:,3))
hold on;grid on;
xlabel('X-Displacement')
ylabel('Y-Displacement')
title('X vs Y Displacement')
end
Matlab error: "Not enough input arguments.
Error in khidonghoc (line 9)
yp(1)= y(2);"
Thanks you very much.
#### 댓글 수: 2
Pass in some arguments! How did you call the function? It expects two input arguments so if you don't pass them in it will give you that error.
Johnny Vendetta 7 Sep 2019
댓글을 달려면 로그인하십시오.
### 채택된 답변
Torsten 3 Sep 2019
function main
tspan=[0 5];
y0=[0;100;0;10];
[t,y]=ode45(@khidonghoc,tspan,y0);
plot(t,y(:,1),t,y(:,3))
hold on;grid on;
xlabel('X-Displacement')
ylabel('Y-Displacement')
title('X vs Y Displacement')
end
function yp=khidonghoc(t,y)
m = 10;
cd = 0.2;
g = 9.81;
w = m*g;
yp = zeros(4,1);
yp(1)= y(2);
yp(2) = ((-cd/m)*y(2)*(y(2)^2+y(4)^2)^(0.5));
yp(3) = y(4);
yp(4) = ((-w/m)-(cd/m)*y(4)*(y(2)^2+y(4)^2)^(0.5));
end
#### 댓글 수: 1
Johnny Vendetta 3 Sep 2019
Thank you so much, Torsten.
댓글을 달려면 로그인하십시오.
### Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by
| 596
| 1,446
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.078125
| 3
|
CC-MAIN-2020-40
|
latest
|
en
| 0.540308
|
http://amathsdictionaryforkids.com/qr/i/integer.html
| 1,685,483,944,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224646144.69/warc/CC-MAIN-20230530194919-20230530224919-00504.warc.gz
| 1,428,614
| 2,886
|
Aa Bb Cc Dd Ee Ff Gg Hh Ii Jj Kk Ll Mm Nn Oo Pp Qq Rr Ss Tt Uu Vv Ww Xx Yy Zz
Ii
integer
• a positive number, a negative number or zero
but not a fraction or a decimal fraction.
EXAMPLES:
Positive + Positive = Positive
Negative + Negative = Negative
Positive + Negative or Negative + Positive
• use the larger number and its sign, subtract
Subtraction (-)
Negative - Positive = Negative
Positive - Negative = Positive
Negative - Negative = Negative + Positive
• use the larger number and its sign, subtract
Multiplication (x)
Positive x Positive = Positive
Negative x Negative = Positive
Negative x Positive = Negative
Positive x Negative = Negative
• change double negatives to a positive
Division (÷)
Positive ÷ Positive = Positive
Negative ÷ Negative = Positive
Negative ÷ Positive = Negative
Positive ÷ Negative = Negative
• change double negatives to a positive
| 224
| 875
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.015625
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.734192
|
https://pt.slideshare.net/sscdotopen/introducing-apache-giraph-for-large-scale-graph-processing
| 1,571,081,425,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986654086.1/warc/CC-MAIN-20191014173924-20191014201424-00472.warc.gz
| 669,582,151
| 41,759
|
Utilizamos seu perfil e dados de atividades no LinkedIn para personalizar e exibir anúncios mais relevantes. Altere suas preferências de anúncios quando desejar.
Próximos SlideShares
Carregando em…5
×
# Introducing Apache Giraph for Large Scale Graph Processing
29.900 visualizações
• Full Name
Comment goes here.
Are you sure you want to Yes No
Tem certeza que deseja Sim Não
Insira sua mensagem aqui
### Introducing Apache Giraph for Large Scale Graph Processing
1. 1. Introducing Apache Giraph for Large Scale Graph Processing Sebastian Schelter PhD student at the Database Systems and Information Management Group of TU Berlin Committer and PMC member at Apache Mahout and Apache Giraph mail ssc@apache.org blog http://ssc.io
2. 2. Graph recapgraph: abstract representation of a set of objects(vertices), where some pairs of these objects areconnected by links (edges), which can be directed orundirectedGraphs can be used to model arbitrary things likeroad networks, social networks, flows of goods, etc.Majority of graph algorithms Bare iterative and traversethe graph in some way A D C
3. 3. Real world graphs are really large!• the World Wide Web has several billion pages with several billion links• Facebook‘s social graph had more than 700 million users and more than 68 billion friendships in 2011• twitter‘s social graph has billions of follower relationships
4. 4. Why not use MapReduce/Hadoop?• Example: PageRank, Google‘s famous algorithm for measuring the authority of a webpage based on the underlying network of hyperlinks• defined recursively: each vertex distributes its authority to its neighbors in equal proportions pj pi dj j ( j , i )
5. 5. Textbook approach to PageRank in MapReduce• PageRank p is the principal eigenvector of the Markov matrix M defined by the transition probabilities between web pages• it can be obtained by iteratively multiplying an initial PageRank vector by M (power iteration) p M p0 k row 1 of M ∙ row 2 of M ∙ pi pi+1 row n of M ∙
6. 6. Drawbacks• Not intuitive: only crazy scientists think in matrices and eigenvectors• Unnecessarily slow: Each iteration is scheduled as separate MapReduce job with lots of overhead – the graph structure is read from disk – the map output is spilled to disk – the intermediary result is written to HDFS• Hard to implement: a join has to be implemented by hand, lots of work, best strategy is data dependent
7. 7. Google Pregel• distributed system especially developed for large scale graph processing• intuitive API that let‘s you ‚think like a vertex‘• Bulk Synchronous Parallel (BSP) as execution model• fault tolerance by checkpointing
8. 8. Bulk Synchronous Parallel (BSP) processorslocal computation superstepcommunicationbarriersynchronization
9. 9. Vertex-centric BSP• each vertex has an id, a value, a list of its adjacent vertex ids and the corresponding edge values• each vertex is invoked in each superstep, can recompute its value and send messages to other vertices, which are delivered over superstep barriers• advanced features : termination votes, combiners, aggregators, topology mutations vertex1 vertex1 vertex1 vertex2 vertex2 vertex2 vertex3 vertex3 vertex3 superstep i superstep i + 1 superstep i + 2
10. 10. Master-slave architecture• vertices are partitioned and assigned to workers – default: hash-partitioning – custom partitioning possible• master assigns and coordinates, while workers execute vertices Master and communicate with each other Worker 1 Worker 2 Worker 3
11. 11. PageRank in Pregelclass PageRankVertex { void compute(Iterator messages) { if (getSuperstep() > 0) { // recompute own PageRank from the neighbors messages pageRank = sum(messages); pj setVertexValue(pageRank); } pi j ( j , i ) dj if (getSuperstep() < k) { // send updated PageRank to each neighbor sendMessageToAllNeighbors(pageRank / getNumOutEdges()); } else { voteToHalt(); // terminate }}}
12. 12. PageRank toy example .17 .33.33 .33 .33 Superstep 0 .17 .17 .17 Input graph .25 .34.17 .50 .34 Superstep 1 A B C .09 .25 .09 .22 .34.25 .43 .34 Superstep 2 .13 .22 .13
13. 13. Cool, where can I download it?• Pregel is proprietary, but: – Apache Giraph is an open source implementation of Pregel – runs on standard Hadoop infrastructure – computation is executed in memory – can be a job in a pipeline (MapReduce, Hive) – uses Apache ZooKeeper for synchronization
| 1,079
| 4,387
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.859375
| 3
|
CC-MAIN-2019-43
|
latest
|
en
| 0.801441
|
http://mathhelpforum.com/discrete-math/126482-supremum-question.html
| 1,480,813,135,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698541142.66/warc/CC-MAIN-20161202170901-00118-ip-10-31-129-80.ec2.internal.warc.gz
| 182,764,583
| 10,206
|
1. ## Supremum Question
Suppose A and B are non-empty sets that are both bounded above, and that for all X in A, there exists a Y in B such that x is less than or equal or y.
a) Prove that sup A is less than or equal to sup B (Need a general proof)
b) Do A = (6/7, 1) and B = {n/(n+1): n in N} satisfy the conditions? Explain.
2. Originally Posted by Janu42
Suppose A and B are non-empty sets that are both bounded above, and that for all X in A, there exists a Y in B such that x is less than or equal or y.
a) Prove that sup A is less than or equal to sup B (Need a general proof)
b) Do A = (6/7, 1) and B = {n/(n+1): n in N} satisfy the conditions? Explain.
We know that $\alpha = \sup (A)\,\& \,\beta = \sup (B)$ both of these exist.
Now suppose that $\beta < \alpha$. That means that $\beta$ is not an upper bound for $A$.
That means $\left( {\exists x \in A} \right)\left[ {\beta < x \leqslant \alpha } \right]$
But by the given $\left( {\exists y\in B} \right)\left[ {x \leqslant y} \right]$
| 322
| 1,001
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625
| 4
|
CC-MAIN-2016-50
|
longest
|
en
| 0.920116
|
http://econlog.econlib.org/archives/2007/04/biases_in_evalu.html
| 1,481,450,161,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698544672.33/warc/CC-MAIN-20161202170904-00216-ip-10-31-129-80.ec2.internal.warc.gz
| 84,998,770
| 11,169
|
Full Site Articles EconLog EconTalk Books Encyclopedia Guides
# Biases in Evaluation Scales
Papers that look interesting... Breast Cancer: a Hansonian Ta...
Elaborating on a point I raised , I am going to make the following conjecture:
In an evaluation scale (e.g, rate this professor on a scale of 1 to 5), the mean evaluation is biased toward the middle.
Is this conjecture false? Is it true, but widely known? If it is true but not published, then someone should formally prove it and submit it to a journal.
When someone is asked to respond to a survey with an answer based on a subjective point scale, the response may not be the person's true best response. The respondent may be reacting to a particular mood, or using a particular interpretation of the question, or making the wrong mark with a pencil.
The average measurement error on points scales is not zero, because the scales are truncated at the endpoints. On a scale of 1 to 5, a person whose true evaluation is 1 can only make an error in one direction--toward a higher number. Similarly, a person whose true evaluation is 5 can only make an error in the other direction--toward a lower number.
If the true group mean is 4.6 out of five, then the measured group mean is likely to be lower, because low-side errors will be included but high-side errors will be truncated. Conversely, if the true group mean is 0.4 out of five, the measured group mean is likely to be higher.
If the scaled evaluation is used as a dependent variable in a regression, the slope of the line will be biased toward zero. For example, suppose that in truth there is a positive relationship between X and Y. If Y's values are biased toward the middle because of measurement error, there will be fewer observations with high-X, high-Y or low-X, low-Y than there would be the case if Y were measured without error.
### Comments and Sharing
CATEGORIES: Economic Methods
COMMENTS (13 to date)
spencer writes:
Businesses use these type of questions all the time and should have very good knowledge of how the answers are biased.
Check with someone in the business or marketing school.
Maniakes writes:
For professor evaluations, a lot of my classmates would give straight "excellent" ratings to all professors regardless of their actual opinion of the professors' performance. I would expect that in cases where the evaluation is perceived to have a real impact on the evaluatee, but there's no cost to the evaluators for overly generous evaluations, the evaluations will be biased towards the top of the scale out of a desire to be nice.
Gary Rogers writes:
I look forward to responses to this question. I do not know what the answer is, but I would venture that the mean is biased toward four simply because most people have no real criteria for making the evaluation and would rather err on the positive side than negative. People like to be generous and a 4 gives them that opportunity while leaving room both above and below for exceptional or poor ratings. To phrase this differently, for most questions people really only have three opinions (excellent, average or poor) and will throw in some generosity to give a rating of three, four or five.
Scott Wood writes:
I don't spend my time working on these problems, so this is a serious question: Is there a practical use to which such information is put for which this observation, if correct, would make any difference?
writes:
[I think maybe Karl Smith meant here:
http://modeledbehavior.blogspot.com/2007/04/evaluation-scales.html
Econlib Editor]
ed writes:
What do you mean "biased?" For a measure to be "biased," there has to be some "true" value that it is biased away from. I'm not sure what it would mean, conceptually, to talk about the "true" values of the responses in these surveys.
Of course if we measure a continuous variable and then censor it on the ends, a formula for the bias is trivial to calculate, given the distribution of the underlying variable.
But in this case, I'm not sure there is any underlying variable that is even defined. If the only variable we are interested in is "how do people answer the question on a five-point scale," then there can't be any bias---the average of the answers in a random sample must be an unbiased estimate of the average answer in the population.
Les writes:
The mean is always biased by outliers. Example: 5 software engineers are at a lunch table. Each earns \$85,000 a year. The mean earnings are \$85,000. Bill Gates sits down at that table. Now mean income is now \$85 million a year. Is the new mean biased? You bet! It is wildly unrepresentative of earnings for 5 of the 6 people at the table.
arthas writes:
This point is more than obvious. When polling institutes ask a question, depending upon how the question is framed, people respond differently.
Secondly, people like to confirm or be nice. They judge what others are expecting as an answer and provide that answer.Thats why most new product launch survey gives a very positive answer while the product launch fails terribly in most cases.
Thirdly, the answer can be culture specific. I had a law professor who would give at max 60 out of 100 for the best student. He reasoned that despite being top of his class at his student years, he never got above sixty and thus giving more than 60 isn't right. Just look at the way professors/ teachers grade their students. I knew a professor in English who would read the first answer and from second answer on award marks based upon the word count.
In my case i always put 3 or 4 out of 5 in any evaluation test where i have no personal feelings as its easy to cross. No need to think
cronbach writes:
Item response theory addresses this and a number of related questions: http://en.wikipedia.org/wiki/Item_response_theory.
The items presented in evaluation scales may be of poor quality.
Fundamentalist writes:
Arnold makes some very good points about rating scales. Another issue is the composition of the audience. The more homogenous the group taking the test is, the more likely it is that you'll get mean scores above or below the median of the rating scale. In my limited teaching experience, I rarely had homogenous groups in which everyone liked me or no one liked me. Usually, a small group liked me and the material very much, a small group hated both, and the rest were asleep. So the average score on the evaluation was in the middle of the scale. Comments reflected the extremes.
bartman writes:
I agree that this is a poor use of the word "biased", especially compared to the sense that the word is used with in econometrics. And what Les is talking about is skewness, not bias.
So, on a 5-point evaluation, does the mean tend to be 3, with the corollary question, is the distribution symmetric, and thus, more or less normal?
My personal experience is no. At three different schools I've worked at the mean is typically closer to 4, with very, very few 1's and 2's. As mentioned above, grade inflation is basically costless, and any prof with half a clue knows how to game the system.
writes:
I work A LOT with survey data based on a 1 to 5 scale, and we rarely see 3 as the mean. It's usually, as another poster said, much closer to 4. In general, different questions will obviously have different mean answers, but from my experience it's usually closer to 4 than to 3.
Yes, you can only make an error in one direction at the edges, but what percentage of people make errors on any given question? That's where the Law of Large Numbers gets involved - the number of errors is usually insignificant relative to any given question, so it can't pull the mean significantly in any direction.
Even though people at the edge can only err in one direction, this only effects the mean if there are differences in the number of people at the edge (or their error rate). If 5 people make an error and write 2 instead of one, and 5 people make an error and write 4 instead of 5, when calculating the mean these errors cancel each other out. The same reason that if 5 people who would have said 3 say 2, and 5 who would have said 3 say 4, the mean doesn't change.
writes:
What do you mean "biased?" For a measure to be "biased," there has to be some "true" value that it is biased away from. I'm not sure what it would mean, conceptually, to talk about the "true" values of the responses in these surveys.
I think Arnold means that the effect of a variable on ranking will be biased downards in maginitude.
The basic assumption that the errors are uncorrelated with the determinate variables is violated.
If for example speaking clearly tended to give someone a high ranking then the errors would be negatively correlated with speaking clearly.
Why?
Because if someone spoke clearly they would tend to get more 5s which have no possibility of a positive error.
If someone did not speak clearly they would tend to get more 1s which have no possibility of a negative error.
Thus the errors move in the opposite direction of the determinate variable. This will tend to downardly bias beta-hat.
The reverse will be true for a determinate variable that is negatively correlated with the ranking.
Comments for this entry have been closed
| 2,001
| 9,258
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.046875
| 3
|
CC-MAIN-2016-50
|
longest
|
en
| 0.915874
|
https://www.numberempire.com/18773
| 1,601,149,808,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400244353.70/warc/CC-MAIN-20200926165308-20200926195308-00340.warc.gz
| 913,938,279
| 6,400
|
Home | Menu | Get Involved | Contact webmaster
Number 18773
eighteen thousand seven hundred seventy three
Properties of the number 18773
Factorization 18773 Divisors 1, 18773 Count of divisors 2 Sum of divisors 18774 Previous integer 18772 Next integer 18774 Is prime? YES (2142nd prime) Previous prime 18757 Next prime 18787 18773rd prime 209621 Is a Fibonacci number? NO Is a Bell number? NO Is a Catalan number? NO Is a factorial? NO Is a regular number? NO Is a perfect number? NO Polygonal number (s < 11)? NO Binary 100100101010101 Octal 44525 Duodecimal aa45 Hexadecimal 4955 Square 352425529 Square root 137.01459776243 Natural logarithm 9.8401749463244 Decimal logarithm 4.2735336801513 Sine -0.91588134165288 Cosine 0.40144908520523 Tangent -2.2814383577052
Number 18773 is pronounced eighteen thousand seven hundred seventy three. Number 18773 is a prime number. The prime number before 18773 is 18757. The prime number after 18773 is 18787. Number 18773 has 2 divisors: 1, 18773. Sum of the divisors is 18774. Number 18773 is not a Fibonacci number. It is not a Bell number. Number 18773 is not a Catalan number. Number 18773 is not a regular number (Hamming number). It is a not factorial of any number. Number 18773 is a deficient number and therefore is not a perfect number. Binary numeral for number 18773 is 100100101010101. Octal numeral is 44525. Duodecimal value is aa45. Hexadecimal representation is 4955. Square of the number 18773 is 352425529. Square root of the number 18773 is 137.01459776243. Natural logarithm of 18773 is 9.8401749463244 Decimal logarithm of the number 18773 is 4.2735336801513 Sine of 18773 is -0.91588134165288. Cosine of the number 18773 is 0.40144908520523. Tangent of the number 18773 is -2.2814383577052
Math tools for your website
Choose language: Deutsch English Español Français Italiano Nederlands Polski Português Русский 中文 日本語 한국어
Number Empire - powerful math tools for everyone | Contact webmaster
By using this website, you signify your acceptance of Terms and Conditions and Privacy Policy.
| 591
| 2,059
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.203125
| 3
|
CC-MAIN-2020-40
|
latest
|
en
| 0.682375
|
https://forum.arduino.cc/t/tmp35-voltage-offset/292296
| 1,675,924,336,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764501407.6/warc/CC-MAIN-20230209045525-20230209075525-00438.warc.gz
| 271,295,312
| 9,106
|
# TMP35 voltage offset?
Hi there,
so I have these 20 tmp35's, and I need to get a full range reading, particularly the low end.
with this setup, we output a voltage between -550mv and 1500mv on a single line relative to ground. So my question is this: how can I get this signal to where a 0-5v arduino analog input pin can read it? what is the simplest/cheapest way to offset these voltages?
Thanks!
There certainly is a way using an OpAmp, but before you buy 20 OpAmp plus other stuff for your 20 sensors , it is simplest to replace the TMP35 with a TMP36 which has the offset "built-in"
That's true, but I have found them to be a bit pricey. Is there no way of doing this conversion with basic components?
Make -Vs the ground, then make a vertuial ground for the sensor sitting Vs above this and finally +Vs to 5V.
Check that the sensor is fine with 5V across it. Make the vertuial ground with two resistors and put a capacitor across each.
hmm, I think I understand you (will draw it all out and do the math later ) but why the capacitor?
so this is what I have so far. Can anyone say whether it looks like this might work?
Yes that is half of what I said. The capacitors are to reduce the impedance of the virtual ground.
However I did say:-
Grumpy_Mike:
Check that the sensor is fine with 5V across it.
I have since checked the data sheet and it says:-
Operates from 4 to 30 volts
And as that circuit has only 2.5V powering it then it will not work, sorry.
Have you got a higher voltage you could use?
You would have to adjust the resistor ratio so that you did not get an output higher than 5V though.
The other option is to generate -5V with a device called a voltage mirror. The ICL7760 is one such device.
He only needs to get the low side about 550mv higher....
So he could provide a virtual ground at, say, +0.8V for it, and he'd still have 4.2v on the sensor, and lowest sensor reading would be 250mv above real ground.
in the drawing, I have listed a 10volt source, with a resistor voltage divider. If I am not mistaken, this should put out +5 and -5 as it is wired there, with the ground in the middle, right? By the way, the data sheet says to choose the resistor based on 50 microamps*-Vs, which I take to mean -5v, so V=IR, 5v=.00005*r, r=100K. is the 5v the value to go off?
As for the capacitors: I'm afraid I know very little about impedance (electronics noob here). It is my understanding that it would be to reduce ac noise? I thought the sensor outputted pure DC, so I'm thinking I still do not understand you. If the capacitors are necessary, what kind do I need to get?
It is my understanding that it would be to reduce ac noise?
Yes
I thought the sensor outputted pure DC
Yes, so what, there is still AC noise about. And the IMPEDANCE is lowered. If you don't understand a word look it up.
If I am not mistaken, this should put out +5 and -5 as it is wired there, with the ground in the middle, right?
What are we talking about here?
It will be -5V ONLY if the Arduino's ground is the virtual ground, but due to the high current you can't make it the Arduino's ground.
So this is what I have tried so far. I hope that unpolished diagram is able to be interpreted. Fortunately, the sensors only take 50 microamps to run, so even running 20 at once is something like a milliamp of power from the arduino. As it stands now, if I hook up a multimeter to my arduino ground, and another to the Vout pin, I get the sensor output plus 5 volts, be the output positive or negative voltage (both are working now).
This is an improvement, as all my output is now positive, but obviously the analog input pins support only up to 5 volts, and depending on the temp, the sensor is outputting between 4.45v and 6.5v. For exampe, at room temperature 22C, I read 5.22 volts. trouble is, the arduino reads up to 5v, so I need a way to read about 3/4 of the output voltage. In order to do that, I attempted to create a voltage divider which would do this. It seems to me like this should work, but I'm getting unpredictable results. For instance, at 5.22v I should be getting 3.912 volts, but I may be reading 3.7v or something.
It seems like the voltage should be divided by the ratio of the voltage divider circuit. I have set this circuit up on 3 breadboards with different resistors, and it isn't. Does anyone know why not, and what could be done to fix it?
Thanks
with different resistors,
What tolerance are the resistors?
I have looked at the datasheet, and it seems you just have to lift all the sensor grounds by e.g. 1volt.
That leaves min 4volt supply (datasheet) for the sensors (Sensors are supplied from the 5volt rail).
That can be done with a resistor divider from the 3.3volt rail.
e.g. 100ohm/220ohm (100ohm to ground). I suppose the 3.3v pin can handle 10mA...
A pulldown resistor (22k for ~50uA) to ground is needed at the output of the sensor to pull it's output below the sensor ground to measure negative C temperatures.
Sensor resolution is 10mv/degreeC. The accuracy measured with the analogue inputs is ~0.5C per digital step.
Zero C will have a digital value of about 205 and 100C about 405.
Leo..
olf2012:
There certainly is a way using an OpAmp, but before you buy 20 OpAmp plus other stuff for your 20 sensors , it is simplest to replace the TMP35 with a TMP36 which has the offset "built-in"
rytcd:
That's true, but I have found them to be a bit pricey. Is there no way of doing this conversion with basic components?
I have looked at pricing from DigiKey, it is exactly the same price for 35 and 36.
\$ 1.2 @ 25 pc
Pelle
Hi,
rytcd, if you reply using REPLY rather than QUICK REPLY, you will find at the bottom of the window a provision to ad files and images as attachents, this will make looking at your images easier.
Thanks Tom.....
They are -+1 percent resistors. I'd expect SOME variance, but not the 10 or 15 degree difference in reading I am seeing
Pelleplutt:
I have looked at pricing from DigiKey, it is exactly the same price for 35 and 36.
\$ 1.2 @ 25 pc
Pelle
I actually got mine 20 for \$15 or so, though if I could do it again I might reconsider. Still, work with what you've got, I guess XD.
@Tom, I actually tried that a couple times. It was not wanting to let me do it. Perhaps there is something I am doin wrong
| 1,633
| 6,308
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.890625
| 3
|
CC-MAIN-2023-06
|
latest
|
en
| 0.959975
|
https://mathematica.stackexchange.com/questions/77662/piecewise-function-output-response-interpretation
| 1,596,635,395,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439735958.84/warc/CC-MAIN-20200805124104-20200805154104-00416.warc.gz
| 409,199,397
| 32,848
|
# Piecewise function output response interpretation
Piecewise[{{t^2, t <= 0}, {t, t >= 0}}]
Outputs:
t^2 t <= 0
t t >= 0
0 True
What does the last line of output 0 True mean? Sometimes it also outputs Indeterminate etc. Can someone describe the behavior, constraints, and limitations of this function?
• see Piecewise >> Details: (1) The Subscript[cond, i] are evaluated in turn, until one of them is found to yield True. _ (2) _If all preceding Subscript[cond, i] yield False, then the Subscript[val, i] corresponding to the first Subscript[cond, i] that yields True is returned as the value of the piecewise function. – kglr Mar 19 '15 at 11:15
• .. (3) Piecewise[{{Subscript[val, 1],Subscript[cond, 1]}, ... },val] uses default value val if none of the Subscript[cond, i] apply. The default for val is 0. and (4) Piecewise[{{Subscript[val, 1],Subscript[cond, 1]},{Subscript[val, 2],Subscript[cond, 2]},...s]}] (i.e, single-argument form with no default value is supplied) uses 0 as the default default-value. – kglr Mar 19 '15 at 11:23
• The function returns 0 if neither t >= 0 nor t <= 0. For example, if you define strange /: strange >= 0 = False; strange /: strange <= 0 = False you get a strange value which is neither >=0 nor <= 0. Consequently, if you evaluate your piecewise function with t=strange, you'll get 0 as result. – celtschk Mar 19 '15 at 16:33
| 414
| 1,382
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.72048
|
https://republicofsouthossetia.org/question/what-s-the-slope-and-y-intercept-of-the-loire-that-is-perpendicular-to-y-2-5-and-passes-through-16355337-53/
| 1,632,074,372,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780056892.13/warc/CC-MAIN-20210919160038-20210919190038-00560.warc.gz
| 522,482,621
| 13,495
|
what’s the slope and y intercept of the loire that is perpendicular to y=-2x-5 and passes through the point (-3,-6)
Question
what’s the slope and y intercept of the loire that is perpendicular to y=-2x-5 and passes through the point (-3,-6)
in progress 0
3 days 2021-09-11T22:23:47+00:00 1 Answer 0
slope: ½
y-intercept: -4½
Step-by-step explanation:
Slope: negative reciprocal of -2
-1/-2 = 1/2
y = ½x + c
-6 = ½(-3) + c
-6 = -3/2 + c
c = -6 + 3/2
c = -9/2 = -4½
| 181
| 476
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2021-39
|
latest
|
en
| 0.826118
|
http://myprimaryclassroom.co.uk/tag/times-tables/
| 1,542,059,603,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039741087.23/warc/CC-MAIN-20181112193627-20181112215627-00343.warc.gz
| 159,407,816
| 15,292
|
## Hard Times : Times Tables Practice
##### Hard Times : Times Tables Practice by transum.org
This is a fun way to bring back pupils focus to the learning of timer tables after a break or simply use for daily practice.
There are five activities to choose from.
Hard Times Plain Game :This is a game for one or many players. Click on a card to see what it contains. Click on a second card and if the two cards make a pair you win them.
Hard Times Matching : Tiles with answers are dragged to their corresponding tables sum.
Hard times Multiple Choice : like the name suggests players drag the tile with the answers to the sum that fits.
Hard Times Tug’o’War: By answering correctly the marker on the tug of war rope will move a little. The objective is to pull the marker past the yellow marker post on your side of the game. Incorrect answers cause the marker to move in the opposite direction. Read the instructions!
Hard Times snap : can be played using keyboard letters or by clicking a button on a tablet.
## Times Tables : Tables Master
##### Tables master Transum.org
Tablesmaster is a great way for students to practise individual times tables.
Played it every day it will help improve numeracy skills, mathematical proficiency and mental strength.
All times tables are included from 2x up to 13x (for the show-offs!)
There is also a printable page on which to record personal best times for completing times tables activities.
This activity is tablet friendly and plays really well on my elderly iPad.
## Arithmetic : A brilliant online times tables activity…
This is one of the best times tables support activities I have seen in a long time.
The game offers three modes on which pupils are offered the opportunity to Multiply, divide or determine factors.
Three Levels of each of the above are on offer: within 6×6, within 9×9 and within 12×12 making this activity suitable for use across the whole primary age range.
Each activity can be attempted against the clock so giving students the opportunity to improve their times and thus their times tables competence. There is also an option to tur off the sound (thank goodness!).
This activity is tablet friendly and works very well on my elderly iPad (double tap required to select a square in the matrix when on iPad)
## “Tables thing” : a times tables practice activity
This activity was originally created so that my class could get competitive with themselves about improving their own times tables knowledge. It provides a written exercise in which pupils complete a 10×10 tables square, the difference being that the numbers along the top and down the left hand side are not sequential, preventing counting on.
By printing off the individual sheets linked below pupils can practise at home using the countdown clock which is also linked. For teachers wishing to use this as a daily morning activity there is also a ‘week to view’ printable.
Tables Thing Original
Numbers 1 – 10
Numbers 1 – 12 excluding 1 and 10
Printable sheet for one week’s practice.
Stopwatch
The way we scored it was that the pupils were given 5 minutes to complete the grid and their score was the number of seconds taken to complete the grid plus the number of blanks and incorrect answers. So a pupil that only filled 84 squares and got 3 wrong would have a score of :
300 seconds allowed + 16 blanks + 3 wrong -= 319
A pupil completing all squares in 275 seconds and making 6 errors would have a score of:
275 seconds used + 6 wrong = 281
This allows pupils to have a number which they can compare to their previous bests and see the improvement they make over time. To do it this way you need to have a countdown clock running so that they can see their time should they finish before the limit.
| 797
| 3,772
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.96875
| 3
|
CC-MAIN-2018-47
|
latest
|
en
| 0.893736
|
https://www.jiskha.com/display.cgi?id=1283742037
| 1,511,529,496,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934808133.70/warc/CC-MAIN-20171124123222-20171124143222-00498.warc.gz
| 834,636,499
| 3,771
|
# chemistry
posted by .
You wish to prepare a 10% CaCl2 solution. Only CaCl2 * 10H2O is availalbe. How many grams of the hydrated compound are required for 250ml of the reagent? Ca=40 Cl=35.5
• chemistry -
If this is 10% by mass (w/w) you can't do it unless you know the density. Is this 10% w/w or 10% w/v? If 10% w/v, which I suspect, convert 10 g CaCl2 to CaCl2.10H2O and that many g in 100 mL will be 10% CaCl2. You need 250 mL; therefore, you will need 2.5 times that amount.
• chemistry -
How do I convert 10 g CaCl2 to CaCl2.10H2O?
## Similar Questions
1. ### chemistry
If 2.1 moles of CaCl2 are dissolved in enough water to make a 0.92-liter solution, what is the resulting molarity?
2. ### chem
158.5 mL of a AgNO3 solution at 5.0M was combined with a 3.5M CaCl2 solution. 110.5g of AgCl was recovered. Given that the CaCl2 is the limiting reagent, how many milliliters of the 3.5M CaCl2 solution was used?
3. ### Chemistry
158.5 mL of a AgNO3 solution at 5.0M was combined with a 3.5M CaCl2 solution. 110.5g of AgCl was recovered. Given that the CaCl2 is the limiting reagent, how many milliliters of the 3.5M CaCl2 solution was used?
4. ### chem
How many grams of CaCl2 would be required to be added to water to prepare 1 Liter (1000ml) of a 1.72% (w/v) aqueous solution of CaCl2?
5. ### Chemistry
how many grams of CaCl2 would be required to be added to water to prepare 1 liter (1000 ml) of a 1.72% (w/v) aqueous solution of CaCl2?
How many grams of CaCl2 would be required to be added to water to prepare 1liter (1000ml) of a 1.72% (w/v) aqueous solution of CaCl2?
7. ### Science
How many grams of CaCl2 are required to prepare 2.00 liters of 7.00 M CaCl2?
8. ### chemistry
how many grams of cacl2 are needed to prepare 6.00L of a 0.400M CaCl2 solution
9. ### Chemistry
How mush CaCl2, in grams, is needed to make 2.0 L of a 3.5M solution?
10. ### chemistry
1. How many grams of calcium chloride (CaCl2) are there in 600.0 mL of an aqueous solution that has a concentration of 0.450 M CaCl2?
More Similar Questions
| 670
| 2,047
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3
| 3
|
CC-MAIN-2017-47
|
latest
|
en
| 0.928135
|
https://forum.allaboutcircuits.com/threads/am-i-doing-this-nodal-analysis-correctly-supernode.121878/
| 1,709,457,017,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947476211.69/warc/CC-MAIN-20240303075134-20240303105134-00724.warc.gz
| 247,926,175
| 21,901
|
# Am I doing this nodal analysis correctly? (SuperNode)
#### Nuno Fernandes
Joined Feb 24, 2016
9
Hello!
I'm having trouble solving a problem with the circuit below. I am supposed to find Va, Vb and Vc through nodal analysis.
I got to these three equations (in the picture, where it says that R3 = 0.333..., it is supposed to be R3 = 1/3):
(Va-Vc)/0.5 + (Va-Vb)/0.5 +2 = 0;
(Vb-Va)/0.5 + Vb/0.25 + (Vc-Va)/0.5 + Vc/(1/3) = 0;
Vc-Va = 3;
I can't get to the correct values I got on the simulation. Is it a problem with the KCL equations? Thank you.
#### StayatHomeElectronics
Joined Sep 25, 2008
1,073
The last equation is not correct. The voltage source is between which two voltages?
#### WBahn
Joined Mar 31, 2012
29,883
Hello!
I'm having trouble solving a problem with the circuit below. I am supposed to find Va, Vb and Vc through nodal analysis.
I got to these three equations (in the picture, where it says that R3 = 0.333..., it is supposed to be R3 = 1/3):
(Va-Vc)/0.5 + (Va-Vb)/0.5 +2 = 0;
(Vb-Va)/0.5 + Vb/0.25 + (Vc-Va)/0.5 + Vc/(1/3) = 0;
Vc-Va = 3;
One habit you want to get into is checking your own set-up equations before you start solving anything. If the set-up isn't correct, then the answers can't be. Remember, the physics and electrical engineering is all in the set-up -- everything else is just math.
You also need to track your units properly -- most mistakes you make will mess up the units allowing you to catch them, but only if the units are being properly used so that they are there to get messed up.
(Va-Vc)/(0.5 Ω) + (Va-Vb)/(0.5 Ω) + 2 A = 0;
(Vb-Va)/(0.5 Ω) + Vb/(0.25 Ω) + (Vc-Va)/(0.5 Ω) + Vc/(1 Ω/3) = 0;
Vc - Va = 3 V;
Go through each node equation (your first two equations) and picture the specific current that each term represents, including which way it is flowing. Ask yourself if those terms properly account for all the current at that node/supernode.
Go through each constraint equation (your third equation) and ask if it represents the correct relationship between the given nodes.
I can't stress enough the importance of checking your set-up equations before proceeding. The whole game is contained in those equations -- those equations ARE both the problem AND the solution. Never assume that you did them correctly. Always check them after you have gotten them written down in order to verify that you didn't mess up -- because you WILL mess up on a regular basis, we all do.
#### RBR1317
Joined Nov 13, 2010
712
This is a good opportunity to explore an alternate method of solution using a super-node. Whenever two node voltages are separated by a known offset, those two nodes can be combined into a super-node. Only one node equation needs to be written for a super-node versus the two equations for the individual nodes. The caveat is that the individual offset node voltages must be used in the node equation as appropriate. To ensure that this is done correctly, mark the offset voltages directly on the diagram.
For this problem I have created the VB super-node, and replaced VC with the VB offset (although it could have been the other way around.) That yields the VA & VB node equations.
#### Nuno Fernandes
Joined Feb 24, 2016
9
This is a good opportunity to explore an alternate method of solution using a super-node. Whenever two node voltages are separated by a known offset, those two nodes can be combined into a super-node. Only one node equation needs to be written for a super-node versus the two equations for the individual nodes. The caveat is that the individual offset node voltages must be used in the node equation as appropriate. To ensure that this is done correctly, mark the offset voltages directly on the diagram.
For this problem I have created the VB super-node, and replaced VC with the VB offset (although it could have been the other way around.) That yields the VA & VB node equations.
View attachment 102427
I'm sorry everyone. It was late and I was really tired and didn't even noticed I had the third equation wrong. Noob mistake Thank you for your reply.
#### anhnha
Joined Apr 19, 2012
905
I'm sorry everyone. It was late and I was really tired and didn't even noticed I had the third equation wrong. Noob mistake Thank you for your reply.
And solve for the equations you should get Va = -(4/7), Vb = -(11/7), Vc = 10/7.
#### WBahn
Joined Mar 31, 2012
29,883
I'm sorry everyone. It was late and I was really tired and didn't even noticed I had the third equation wrong. Noob mistake Thank you for your reply.
Which underscores the value and importance of getting in the habit of always (not sometimes, but always) verifying your set-up equations before proceeding. It actually wasn't a noob mistake -- there will always be times when you are tired, or hungry, or hurried, or stressed, or just plain ole a little sloppy. That will be true throughout your career. So you want to set up best practices that, once they become engrained habit, will let you automatically catch most of your mistakes regardless of why they got made.
| 1,296
| 5,042
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.328125
| 3
|
CC-MAIN-2024-10
|
latest
|
en
| 0.944988
|
https://notebooks-prod.herokuapp.com/previews?key=5150fd5e153d46083d0002df%2F6fec8792-f359-439c-b4e7-9457921b07ef%2Fpreview.html&vf=true
| 1,591,510,455,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590348523564.99/warc/CC-MAIN-20200607044626-20200607074626-00511.warc.gz
| 448,137,322
| 46,611
|
Notebook
# Quantopian's slippage model for futures¶
This notebook is an overview of Quantopian's slippage model for futures, VolatilityVolumeShare. If you aren't familiar with what slippage is, consider a simple example: buying futures contracts for coffee. We want to buy five contracts at \$500 apiece. We buy the first four, but by the time we buy the fifth, the price of the coffee contract has increased to \$503. This change in price is what is known as slippage. It will typically work against us — our own buys in a market increase demand, which in-turn increases price.
Our slippage model for futures should, within the backtester, slightly increase the price on a buy and slightly decrease the price on a sell. We can do this similarly to how we apply slippage to stocks. If $MI$ is our market impact, $P_0$ is the original price of the future, and $P$ is the price with slippage applied, then
#### $P = P_0 \left(1 + MI\right)$¶
$MI$ will be positive for buys and negative for sells.
To model slippage here, we need to model $MI$. One way to model $MI$ is with
### $MI = \eta \sigma \sqrt{\psi}$¶
Where $\sigma$ is volatility (in this case, 20-day annualized daily volatility), $\psi = \frac{our\ volume\ traded\ in\ bar}{20-day\ ADV}$, and $\eta$ is a constant we fit for every continuous future.
To fit the constant $\eta$ for each future, we first compute it once per month for the asset. We do this by looking at the trailing 60 days of price and volume minute bars, and regressing on
### $\left|r\right| = \eta\sigma\sqrt{J}$¶
Where $r$ is the percentage change of the asset's price in each bar (which we take the absolute value of) and $J$ is $\frac{market\ volume\ traded\ in\ bar}{20-day\ ADV}$ for every bar. This method is known as Kyle's Lambda.
Once we have a value of $\eta$ for each month in a certain time frame (5 years maximum for each future, in this case), we average these values together to get the final value of $\eta$ we will use for each continuous future. This notebook is meant to calculate these $\eta$ values. It can also be run without the volatility ($\sigma$) term.
### Examples¶
• Say we're trying to buy 100 front contracts of ES, the E-Mini S&P 500, in one minute. If E-Mini is trading at \$2000, with an ADV of 1.4 million contracts/day and a volatility of 9%. Eta for ES is 0.047. Our market impact is then$MI = 0.047 \cdot 0.09 \cdot \sqrt{\frac{100}{1,400,000}} = 0.00003575 = 0.3575 \ \text{bps}$. (bps, or [basis points](https://en.wikipedia.org/wiki/Basis_point), is one-hundredth of one percent). Therefore, the price we would buy E-Mini at here, according to our slippage model, is$ \$2000 \cdot \left(1 + 0.00003575\right) = \$2000.0715 $. • In this example, we'll pretend to buy 50 contracts of CL, the crude oil future, in one minute. Pretend CL is trading at \$60 with an ADV of 400,000 contracts and a volatility of 23%. Eta is 0.049. In this case, our market impact is $MI = 0.049 \cdot 0.23 \cdot \sqrt{\frac{50}{400,000}} = 1.26 \ \text{bps}$. Our buy price is then $\$50 \cdot \left(1 + 0.000126\right) = \$50.0063$.
In [1]:
from __future__ import division
import numpy as np
import pandas as pd
import empyrical
from quantopian.research.experimental import continuous_future, history
# This can take a while to run, so we'll only do a few here
root_symbols = ['HG', 'HO', 'GC'] # 'HU', 'QM', 'PA', 'XG', 'PL', 'CL',
#'QG', 'YS', 'SV', 'NG', 'XB', 'EU', 'NZ', 'JY', 'EC',
#'ME', 'AD', 'SF', 'EL', 'EE', 'BP', 'JE', 'CD', 'LB',
#'WC', 'ET', 'SM', 'FC', 'CM', 'CN', 'MW', 'PB', 'LC',
#'SB', 'MS', 'OA', 'SY', 'LH', 'RR', 'BO', 'UB', 'FS',
#'TB', 'TY', 'FV', 'TS', 'TU', 'FF', 'MB', 'FI', 'TN',
#'US', 'ED', 'MI', 'NK', 'MD', 'RM', 'DJ', 'MG', 'NQ',
#'ER', 'BD', 'EI', 'ES', 'VX', 'SP', 'ND', 'YM', 'AI']
# Only fit for some data. Before '05, there's not a lot of electronic trades.
max_days = pd.Timedelta(days=365 * 5.0)
# Frequency. 'D' for calculating eta once/day, 'M' for once/month.
eta_freq = 'M'
# If False, fit abs(bar_returns) = eta*sqrt(pct_of_ADV)
# If True, fit abs(bar_returns) = eta*volatility*sqrt(pct_of_ADV)
include_volatility = True
In [2]:
# Computes eta for a given root symbol, as described above
# This can take a while to run
def compute_eta(symbol):
cf = continuous_future(symbol, offset=0, roll='volume', adjustment='mul')
cf_end_date = cf.end_date - pd.Timedelta(days=1)
cf_start_date = max(cf_end_date - max_days,
cf.start_date,
pd.Timestamp('2002-02-01', tz='UTC'))
if cf_start_date > cf_end_date:
return None
market_data = history(
cf,
fields=['price', 'volume'], # We want the forward-filled price
frequency='minute',
start_date=cf_start_date + pd.Timedelta(hours=12),
end_date=cf_end_date + pd.Timedelta(hours=12)
)
market_data.index = market_data.index.tz_convert('EST')
# Get daily volume & close
daily_volume = market_data.volume.resample('1D').sum().dropna()
daily_close = market_data.price.resample('1D').last().dropna()
daily_returns = daily_close.pct_change().dropna()
# Calculate 20 day volatility and ADV. Only include previous days.
vol_20d = daily_returns.rolling(21).agg(lambda x: empyrical.annual_volatility(x[:-1]))
# Calculate regression variables
market_data['returns'] = market_data.price.pct_change()
market_data['abs_returns'] = market_data.returns.abs()
market_data['poadv'] = market_data.apply(lambda x: x.volume / daily_volume[x.name.normalize()]
if x.name.normalize() in daily_volume else np.nan, axis=1)
market_data['vol_20d'] = market_data.apply(lambda x: vol_20d[x.name.normalize()]
if x.name.normalize() in vol_20d else np.nan, axis=1)
# Delete rows where the previous row wasn't 1 minute earlier
# We only want to capture minute-to-minute price changes
index_series = market_data.index.to_series()
rows_to_drop = index_series[index_series.diff() != pd.Timedelta(minutes=1)].index
market_data = market_data.drop(rows_to_drop)
# If volume is 0, the price shouldn't change
market_data = market_data[market_data.volume > 0]
if include_volatility:
else:
sample_size = len(market_data)
print symbol, 'samples:', sample_size
if sample_size < 2: # Confirm the sample size is large enough
return None
etas = []
days = pd.date_range(cf_start_date + pd.Timedelta(days=60), cf_end_date, freq=eta_freq)
for day in days:
sample_market_data = market_data[day - pd.Timedelta(days=60):day]
if len(sample_market_data) < 2:
return None
if include_volatility:
y = sample_market_data.abs_returns.values
x = x[:,np.newaxis]
slope, _, _, _ = np.linalg.lstsq(x, y)
else:
y = sample_market_data.abs_returns.values
x = x[:,np.newaxis]
slope, _, _, _ = np.linalg.lstsq(x, y)
etas.append(slope)
return etas, sample_size
In [3]:
all_etas = {}
all_sample_sizes = {}
# Iterate through each future
for symbol in root_symbols:
try:
res = compute_eta(symbol)
if res is not None:
all_etas[symbol] = res[0]
all_sample_sizes[symbol] = res[1]
except Exception as e:
print symbol, ':', e
# Mean eta for each symbol
mean_eta = {symbol: np.mean(etas) for symbol, etas in all_etas.iteritems()}
HG samples: 1546475
HO samples: 1086430
GC samples: 1724718
In [4]:
# Only take etas with a large enough sample size
# Also compute a sample-size weighted mean eta, as a fallback for thinly-traded names
min_sample_size = 10000
clean_etas = {}
weighted_sum = 0
total_weight = 0
for symbol, eta in mean_eta.iteritems():
ss = all_sample_sizes[symbol]
if eta < 1: # The result must be reasonable
weighted_sum += eta * ss
total_weight += ss
if ss > min_sample_size:
clean_etas[symbol] = eta
grand_mean_eta = weighted_sum / total_weight
print 'Symbol etas:', clean_etas
print 'Default eta: ', grand_mean_eta
Symbol etas: {'GC': 0.04891437779901682, 'HG': 0.052224109334103576, 'HO': 0.045047038237878125}
Default eta: 0.0491247730894
In [ ]:
| 2,270
| 7,753
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.28125
| 3
|
CC-MAIN-2020-24
|
latest
|
en
| 0.909294
|
https://www.cnblogs.com/birchtree/p/11715607.html
| 1,638,805,309,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964363301.3/warc/CC-MAIN-20211206133552-20211206163552-00442.warc.gz
| 774,029,489
| 8,248
|
# [BZOJ 3771] Triple(FFT+容斥原理+生成函数)
## [BZOJ 3771] Triple(FFT+生成函数)
### 分析
$B(x)=\sum_{i=1}^{n} x^{2w_i}$
$C(x)=\sum_{i=1}^{n} x^{3w_i}$
#### 取3个不同物品的情况
$\frac{A^3(x)-3A(x)B(x)+2C(x)}{6}$
#### 取2个不同物品的情况
$\frac{A^2(x)-B(x))}{2}$
#### 取1个不同物品的方案
$\frac{A^3(x)-3A(x)B(x)+2C(x)}{6}+\frac{A^2(x)-B(x))}{2}+A(x)$
### 代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define maxn 400000
using namespace std;
typedef long long ll;
const double pi=acos(-1.0);
struct com{
double real;
double imag;
com(){
}
com(double _real,double _imag){
real=_real;
imag=_imag;
}
com(double x){
real=x;
imag=0;
}
void operator = (const com x){
this->real=x.real;
this->imag=x.imag;
}
void operator = (const double x){
this->real=x;
this->imag=0;
}
friend com operator + (com p,com q){
return com(p.real+q.real,p.imag+q.imag);
}
friend com operator + (com p,double q){
return com(p.real+q,p.imag);
}
friend com operator - (com p,com q){
return com(p.real-q.real,p.imag-q.imag);
}
friend com operator - (com p,double q){
return com(p.real-q,p.imag);
}
friend com operator * (com p,com q){
return com(p.real*q.real-p.imag*q.imag,p.real*q.imag+p.imag*q.real);
}
friend com operator * (com p,double q){
return com(p.real*q,p.imag*q);
}
friend com operator / (com p,double q){
return com(p.real/q,p.imag/q);
}
void print(){
printf("%lf + %lf i ",real,imag);
}
};
void fft(com *x,int n,int type){
static int rev[maxn+5];
int tn=1,k=0;
while(tn<n){
k++;
tn*=2;
}
for(int i=0;i<n;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(k-1));
for(int i=0;i<n;i++) if(i<rev[i]) swap(x[i],x[rev[i]]);
for(int len=1;len<n;len*=2){
int sz=len*2;
com wn1=com(cos(2*pi/sz),type*sin(2*pi/sz));
for(int l=0;l<n;l+=sz){
int r=l+len-1;
com wnk=1;
for(int i=l;i<=r;i++){
com tmp=x[i+len];
x[i+len]=x[i]-wnk*tmp;
x[i]=x[i]+wnk*tmp;
wnk=wnk*wn1;
}
}
}
if(type==-1){
for(int i=0;i<n;i++) x[i].real/=n;
}
}
void mul(com *a,com *b,com *ans,int n){
// fft(a,n,1);
// fft(b,n,1);
//避免多次fft
for(int i=0;i<n;i++) ans[i]=a[i]*b[i];
fft(ans,n,-1);
}
int n;
int val[maxn+5];
com a[maxn+5],b[maxn+5],c[maxn+5];
com ans[maxn+5];
int main(){
scanf("%d",&n);
int mv=0;
for(int i=1;i<=n;i++){
scanf("%d",&val[i]);
a[val[i]]=a[val[i]]+1;
b[val[i]*2]=b[val[i]*2]+1;
c[val[i]*3]=c[val[i]*3]+1;
mv=max(mv,val[i]);
}
int tn=1,k=0;
while(tn<mv*3){
k++;
tn*=2;
}
fft(a,tn,1);
fft(b,tn,1);
fft(c,tn,1);
for(int i=0;i<tn;i++){
ans[i]=(a[i]*a[i]*a[i]-3*a[i]*b[i]+2*c[i])/6+(a[i]*a[i]-b[i])/2+a[i];
}
fft(ans,tn,-1);
for(int i=0;i<=mv*3;i++){
if(ll(ans[i].real+0.5)){
printf("%d %lld\n",i,ll(ans[i].real+0.5));
}
}
}
posted @ 2019-10-21 19:48 birchtree 阅读(227) 评论(0编辑 收藏 举报
| 1,074
| 2,627
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.421875
| 3
|
CC-MAIN-2021-49
|
latest
|
en
| 0.122697
|
https://www.sarthaks.com/781564/find-capacitance-capacitor-voltage-across-volt-current-passing-through-10-ampere-50hz
| 1,618,117,086,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038060927.2/warc/CC-MAIN-20210411030031-20210411060031-00467.warc.gz
| 1,110,105,545
| 13,519
|
# Find the capacitance of a capacitor if the voltage across it is 100 volt and the current passing through it is 10∠-90 ° ampere (f = 50Hz)
56 views
edited
Find the capacitance of a capacitor if the voltage across it is 100 volt and the current passing through it is 10∠-90 ° ampere (f = 50Hz)
+1 vote
by (67.2k points)
We have V=100 volt
| 101
| 343
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2021-17
|
latest
|
en
| 0.832211
|
https://cs.nyu.edu/pipermail/fom/2009-March/013461.html
| 1,685,743,966,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224648858.14/warc/CC-MAIN-20230602204755-20230602234755-00532.warc.gz
| 217,447,839
| 3,230
|
# [FOM] The boundary of objective mathematics
Paul Budnik paul at mtnmath.com
Mon Mar 9 20:50:58 EDT 2009
```Henrik Nordmark wrote:
> ...
>
>
> What I find interesting is that you seem to suggest that mathematical
> statements can be partitioned into objective statements and non-
> objective statements. Most philosophical camps attempt to have one
> overarching status for all mathematical statements. You seem to be a
> realist for a certain subclass of statements and an anti-realist for
> the rest. Regardless of where exactly where one draws a boundary, I am
> wondering what kinds of problems does drawing a boundary generate.
> Clearly, having one status for all mathematical statements is more
> elegant but that hardly seems like a serious philosophical objection
>
The natural instinct is to make a formal system for mathematics as
powerful as possible. This has led to inconsistencies in the past but it
can also lead to a formal system that is consistent but incorrect. For
example it might be consistent but not omega consistent. I suspect that
ZF does not violate any generalization of omega consistency but I still
think it is too powerful and thus in some sense incorrect.
It is incorrect in the sense of treating some relative statements as if
they were absolute because, in some respects, it treats infinite objects
as if they had something like a physical existence. This begins with the
power set axiom. The set of all subsets of the integers definable within
ZF is well defined. One can enumerate all the statements in ZF that
provably define a particular real. But this is not what mathematicians
want the set of all subsets of the integers to mean. However there is no
way to enforce what they want it to mean and when they argue as if what
they mean is objective they are treating a relative statement as if it
were absolute.
I an dividing mathematics into absolute objective statements and
relative ones that are sometimes treated as if they were absolute.
> However, I suspect that your position probably does have some problems
> with it. For example, you seem to accept the existence of real numbers
> and natural numbers, so these would be amongst your objective and true
> statements. But then, one can ask whether there is an infinite subset
> of the reals which is not in a bijection to either the reals nor the
> natural numbers. Since we are dealing with real objects, presumably
> this question must have a definite and objective answer, but this
> mathematical statement.
>
Real numbers are problematic because they seem so natural and obvious. I
think we can talk about an arbitrary integer parameter for a recursive
process. And I think we can talk about an arbitrary infinite sequence of
integer parameters. This all makes sense to me in an always finite but
potentially infinite universe. Asking what a TM will do for all possible
infinite sequences of inputs make sense because every event that
determines that can be recursively enumerated. However I think this
question is a human creation. It is not a question about something that
can ever exist. It only asks about the unbounded future of a recursive
process in a potentially infinite universe.
In contrast the CH asks about what infinite sets exist in an absolute
sense. The question is meaningful relative to the sets provably
definable in a formal system but it has no absolute meaning because
infinite sets are human creations and abstractions.
> In all fairness, I do not know your exact position well enough to
> assess whether the example I just gave would actually be problematic
> for the stance you are trying to take. However, it does make me wonder
> if one can develop a philosophical stance with a realist/anti-realist
> dichotomy built into it that is not exceedingly problematic.
>
The difference I want to make is between an an arbitrary path that is
followed by a recursive process in a potentially infinite universe and a
completed infinite set. This is a distinction that goes back at least
to Aristotle. My position is perhaps close to constructivists, but I do
not demand a constructive proof of a statement. I only demand a
constructive proof that all the events that determine the statement are
themselves determined by finite events.
Paul Budnik
www.mtnmath.com
```
| 906
| 4,296
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.965131
|
https://stats.stackexchange.com/questions/58381/mathematically-modeling-neural-networks-as-graphical-models
| 1,702,284,588,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-50/segments/1700679103810.88/warc/CC-MAIN-20231211080606-20231211110606-00361.warc.gz
| 600,530,708
| 45,783
|
# Mathematically modeling neural networks as graphical models
I am struggling to make the mathematical connection between a neural network and a graphical model.
In graphical models the idea is simple: the probability distribution factorizes according to the cliques in the graph, with the potentials usually being of the exponential family.
Is there an equivalent reasoning for a neural network? Can one express the probability distribution over the units (variables) in a Restricted Boltzmann machine or a CNN as a function of their energy, or the product of the energies between units?
Also, is the probability distribution modelled by an RBM or Deep belief network (e.g. with CNNs) of the exponential family?
I am hoping to find a text that formalizes the connection between these modern types of neural networks and statistics in the same way that Jordan & Wainwright did for graphical models with their Graphical Models, Exponential Families and Variational Inference. Any pointers would be great.
• IM (hater's) O the core problem here is that neural networks are not really networks; they practically have a fixed topology and thus have a minor chance to store any information inside it.
– user88
May 7, 2013 at 15:35
• Have you seen this recent post? May 7, 2013 at 16:27
• @jerad Thanks, I hadn't read that post. My question is not so much on how to combine these models (e.g. such as when Yann says "using deep nets as factors in an MRF"), but more about how to look at a deep net as a probabilistic factor graph. When Yann LeCun's says "of course deep Boltzmann Machines are a form of probabilistic factor graph themselves", I am interested in seeing that connection mathematically. May 7, 2013 at 16:56
• @mbq, we've seen some forms of hidden layer component information storage, eg https://distill.pub/2017/feature-visualization/ (How neural networks build up their understanding of images), in that a complex image has component objects represented by hidden layer nodes. The weights can 'alter' the 'topology' in a non-discrete fashion. Although I have not seen it, some methods could include shrinkage factors to remove edges and therefore change the original topology
– Vass
Mar 23, 2019 at 17:48
Another good introduction on the subject is the CSC321 course at the University of Toronto, and the neuralnets-2012-001 course on Coursera, both taught by Geoffrey Hinton.
From the video on Belief Nets:
## Graphical models
Early graphical models used experts to define the graph structure and the conditional probabilities. The graphs were sparsely connected, and the focus was on performing correct inference, and not on learning (the knowledge came from the experts).
## Neural networks
For neural nets, learning was central. Hard-wiring the knowledge was not cool (OK, maybe a little bit). Learning came from learning the training data, not from experts. Neural networks did not aim for interpretability of sparse connectivity to make inference easy. Nevertheless, there are neural network versions of belief nets.
My understanding is that belief nets are usually too densely connected, and their cliques are too large, to be interpretable. Belief nets use the sigmoid function to integrate inputs, while continuous graphical models typically use the Gaussian function. The sigmoid makes the network easier to train, but it is more difficult to interpret in terms of the probability. I believe both are in the exponential family.
I am far from an expert on this, but the lecture notes and videos are a great resource.
• Welcome to the site. We are trying to build a permanent repository of high-quality statistical information in the form of questions & answers. Thus, we're wary of link-only answers, due to linkrot. Can you post a full citation & a summary of the information at the link, in case it goes dead? Apr 27, 2015 at 14:32
• This is really nice. Thank you for adding this information & welcome to CV. Apr 27, 2015 at 16:43
• I have to point out that the information in the first half of your answer is not quite accurate, which I guess is implied by the use of "early graphical models" (should be "very very early"). For a very long time, graphical models have been used to learn all aspects of its architecture the same way as neural networks have. But your later suggestion on sigmoids taking the place of gaussians in factor graphs is interesting! Nov 20, 2015 at 19:21
Radford Neal has done a good bit of work in this area that might interest you, including some direct work in equating Bayesian graphical models with neural networks. (His dissertation was apparently on this specific topic.)
I'm not familiar enough with this work to provide an intelligent summary, but I wanted to give you the pointer in case you find it helpful.
• From what I understand from works of Neal, Mackay etc, they are using Bayesian Optimization where the parameters to optimize over are the neural weights and biases, even going to show that the L2 normalization of neural networks can be seen as a Gaussian prior over the weights. That program has been continued to include number of hidden layers, neurons within each layer etc among the optimization variables. Nov 20, 2015 at 19:11
• But this is different from what the OP asked because designing the architecture of the neural network to tryout in the next run is just one special case of experimental design using Bayesian models as a hyper-design engine. I think what the OP asked for was a mapping between neural network and bayesian modeling, at the "same level". Nov 20, 2015 at 19:14
This may be a old thread, but still a relevant question.
The most prominent example of the connections between Neural Networks (NN) and Probabilistic Graphical Models (PGM) is the one between Boltzmann Machines (and its variations like Restricted BM, Deep BM etc) and undirected PGMs of Markov Random Field.
Similarly, Belief Networks (and it's variations like Deep BN etc) are a type of directed PGMs of Bayesian graphs
For more, see:
1. Yann Lecun, "A tutorial on energy-based learning" (2006)
2. Yoshua Bengio, Ian Goodfellow and Aaron Courville, "Deep Learning", Ch 16 & 20 (book in preparation, at the time of writing this)
| 1,369
| 6,212
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2023-50
|
longest
|
en
| 0.9421
|
https://byjus.com/question-answer/a-man-running-on-a-horizontal-road-at-8-text-ms-1-finds-rain-falling-8/
| 1,679,673,956,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00637.warc.gz
| 190,058,221
| 28,528
|
Question
# A man running on a horizontal road at 8 ms−1 finds rain falling vertically. If he increases his speed to 12 ms−1, he finds that drops make 30∘ angle with the vertical. Find the velocity of rain with respect to the road.
A
47 ms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
82 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
73 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
8 ms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is A 4√7 ms−1Let velocity of rain Vr=Vrx^i+Vry^j Velocity of rain wrt man, Vrm=(Vrx−Vm)^i+Vry^j case (i) Vrx−Vm=0⇒vrx=8 ms−1 case (ii) tan30=Vrx−VmVry=8−12Vry Vry=−4√3 ms−1 Vr=8^i−4√3^j=4√22+3=4√7 ms−1
Suggest Corrections
0
Related Videos
Relative Motion in 2D
PHYSICS
Watch in App
| 317
| 859
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.546875
| 4
|
CC-MAIN-2023-14
|
latest
|
en
| 0.738341
|
https://tutorstips.com/question-no-40-chapter-no-13-usha-publication-11-class/
| 1,632,140,836,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00529.warc.gz
| 645,153,336
| 20,144
|
# Question No 40 Chapter No 13 – USHA Publication 11 Class
Question No 40 Chapter No 13
Creation of Provision for Depreciation Account
40. The following balances appear in the books of Jain Mills:
April.1 2018 Machinery Account Rs 80,000 Provision for Depreciation Rs 30,000
On April 1, 2018 they decided to sell machinery for Rs 8,350. this machine was purchased for Rs 16,000 on 1st April 2014.
You are required to prepare Machinery Account and provision for depreciation Account for the year 2018-19, assuming the firm has been charging depreciation at 10% p.a. on original cost method
The solution of Question No 40 Chapter No 13:-
Dr. Machinery A/c Cr. Dat Particulars J.F. Amount Date Particulars J.F. Amount 01/03/18 To Balance b/f 80,000 01/03/18 By Machinery Disposal A/c 16,000 31/03/19 By Balance C/d 64,000 80,000 80,000
Dr. Provision for Depreciation A/c Cr. Date Particulars J.F. Amount Date Particulars J.F. Amount 01/03/18 To Machinery Disposal A/c 6,400 01/03/18 By Balance b/f 30,000 31/03/19 To Balance c/d 30,000 31/03/19 By Depreciation A/c 6,400 36,400 36,400
Dr. Machinery Disposal A/c Cr. Date Particulars J.F. Amount Date Particulars J.F. Amount 01/03/18 By Balance b/f 8,350 01/03/18 By Provision for Dep. A/c 6,400 01/03/18 By Profit/Loss A/c 1,250 31/03/19 To Balance c/d 16,000 16,000 16,000
Working note:-
Statement Showing profit or loss on the sale of Machinery Particulars Amount Purchase value of Equipment as on 1st Apr, 2014 16,000 Less: – Amount of Depreciation charged on the year 2014-15 16,000 *10%* 12/12 1,600 Amount of Depreciation charged on the year 2015-16 16,000 *10%* 12/12 1,600 Amount of Depreciation charged on the year 2016-17 16,000 *10%* 12/12 1,600 Amount of Depreciation charged on the year 2017-18 16,000 *10%* 12/12 1,600 Book value of the asset as on 1st March 2018 9,600 Sale Price of Machinery 8,350 Loss on the sale of the asset 1,250
\
Depreciation | Meaning | Methods | Examples
Comment if you have any question.
Also, Check out the solved question of all Chapters: –
Chapter No. 1 – Introduction
Chapter No. 2 – Theory Base of Accounting
Chapter No. 3 – Vouchers and transactions
Chapter No. 4 – Journal
Chapter No. 5 – Ledger
Chapter No. 6 – Cash Book
Chapter No. 7 – Other Subsidiary Books
Chapter No. 8 – Journal Proper
Chapter No. 9 – Trial Balance
Chapter No. 10 – Bank Reconciliation Statement
Chapter No. 11 – Depreciation
Chapter No. 12 – Provisions and Reserves
Chapter No. 13 – Bills of Exchange
Chapter No. 14 – Rectification of Errors
Chapter No. 15 – Financial Statements – (Without Adjustments)
Chapter No. 16 – Financial Statements – (With Adjustments)
| 828
| 2,668
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.734375
| 3
|
CC-MAIN-2021-39
|
latest
|
en
| 0.800284
|
https://clarksperformancediesel.com/internal-combustion-engine/your-question-what-is-zero-in-an-induction-motor.html
| 1,660,238,572,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882571483.70/warc/CC-MAIN-20220811164257-20220811194257-00007.warc.gz
| 184,841,684
| 18,403
|
# Your question: What is zero in an induction motor?
Contents
## What happens when slip is zero in induction motor?
Zero slip means that rotor speed is equal to synchronously rotating magnetic flux. … Hence, no emf will be generated in rotor coils to produce rotor current. This means no electromagnetic torque will be produced. Induction motor will not work.
## What is speed in induction motor?
Synchronous and full load speed of amplitude current (AC) induction motors
Speed (rpm)
Number of Poles Frequency (Hz, cycles/sec)
2 3600 2900
4 1800 1450
6 1200 960
## Why there is no slip in synchronous motor?
Because the rotor turns at the same speed as synchronous speed (speed of the rotating magnetic field), there is no slip. The speed of rotation of the motor is constant in a synchronous motor, and does not vary with load, as in an induction motor.
## What causes slip in induction motor?
“Slip” in an AC induction motor is defined as:
As the speed of the rotor drops below the stator speed, or synchronous speed, the rotation rate of the magnetic field in the rotor increases, inducing more current in the rotor’s windings and creating more torque.
## Is 3 phase induction motor is self starting?
Three-phase induction motor is self-starting, because winding displacement is 120 degrees for each phase and supply also has 120 phase shift for 3-phase. It results in a unidirectional rotating magnetic field is developed in air gap which causes 3-phase induction motor to self-start.
IT IS INTERESTING: Frequent question: What is an electric motor quizlet?
## What is synchronous speed?
Synchronous speed is a significant parameter for the rotating magnetic field-type AC motor. It is determined by the frequency and the number of magnetic poles. A collective name for the motors that run at the synchronous speed is the synchronous motor.
## Can slip be negative?
If slip is negative, the “input power” to the electrical terminals will be negative, implying that power is flowing out of the electrical terminals. Rotational losses are 2450W.
## What is rpm in motor?
CARS.COM — RPM stands for revolutions per minute, and it’s used as a measure of how fast any machine is operating at a given time. In cars, rpm measures how many times the engine’s crankshaft makes one full rotation every minute, and along with it, how many times each piston goes up and down in its cylinder.
## What is induction motor full load?
Rated full-load speed — This is the motor’s approximate speed under full-load conditions, when voltage and frequency are at the rated values. … On standard induction motors, the full-load speed is typically 96% to 99% of the no-load speed.
| 579
| 2,683
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3
| 3
|
CC-MAIN-2022-33
|
latest
|
en
| 0.885756
|
http://kchemistry.com/thermo_notes.htm
| 1,621,245,860,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243992159.64/warc/CC-MAIN-20210517084550-20210517114550-00362.warc.gz
| 27,903,117
| 2,830
|
When heating up ice, the first thing we need to notice is that the curve is not a straight line, but has 5 straight segments. Each segment is " ruled " by a different equation. To calculate the total energy change (delta H) of a change, calculate the energy of each segment and add them together. 1. In this region we are heating ice. The formula for how energy is needed is: ΔH = m * ΔT*Cp. Where ΔH is the change in energy needed, m is the mass of ice being warmed or cooled, ΔT is the temperature change (Ending T- Beginning T), and Cp is the specific heat of the ice (0.5 cal/g °C) 2. In this region a curious thing happens, the temperature neither goes up or down, all the heat energy is used in melting or freezing. The formula for how energy is needed is: ΔH = m * Hf Where ΔH is the change in energy needed, m is the mass of ice melted or the mass of water being frozen, Hf is the heat of fusion of ice (80 cal/g) 3. In this region we are warming water, with no phase change so the equation is: ΔH = m * ΔT*Cp. Where ΔH is the change in energy needed, m is the mass of ice being warmed or cooled, ΔT is the temperature change (Ending T- Beginning T), and Cp is the specific heat of the water (1.0 cal/g °C) 4. In this region another phase change occurs.Water is converted to steam (or steam to water) without any temperature rise. The formula for how energy is needed is: ΔH = m * Hv Where ΔH is the change in energy needed, m is the mass of water vaporized or the mass of steam being condensed, Hv is the heat of vaporization of water (a huge 540 cal/g) 5. In this region we are heating steam. The formula for how energy is needed is: ΔH = m * ΔT*Cp. Where ΔH is the change in energy needed, m is the mass of steam being warmed or cooled, ΔT is the temperature change (Ending T- Beginning T), and Cp is the specific heat of the steam (0.5 cal/g °C) To solve numerical problems plot the beginning point and ending point on the graph and break the heating into small segments (up to 5) calculate the energy change for each step and add them together for the total energy change of the total "trip". Calculate using only positive numbers and assign a sign at the very end: left and down represents - calories; up and right is + calories. In Mr. K's class we will use calories so we can relate to food calories (kilocalories), and to relate to the rest of the world we will use joules (pronounced jewels). 1 calorie [thermochemical] = 4.184 joule 1 food Calorie = 1000 chemistry calories = 1 kcal.Note: physicists (and some chemists) use Q to denote energy. If we only work at constant pressure (usually atmospheric) then Q = ΔH. Their terminology is a little more universal it allows for PdV work. Ours is a little more simple and ΔH reminds me of heat.
| 715
| 2,791
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.96875
| 4
|
CC-MAIN-2021-21
|
longest
|
en
| 0.906939
|
https://encyclopedia2.thefreedictionary.com/tighten+the+screw
| 1,607,172,101,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141747774.97/warc/CC-MAIN-20201205104937-20201205134937-00673.warc.gz
| 279,602,996
| 15,858
|
screw
(redirected from tighten the screw)
Also found in: Dictionary, Thesaurus, Medical, Idioms.
screw,
simple machinemachine,
arrangement of moving and stationary mechanical parts used to perform some useful work or to provide transportation. From a historical perspective, many of the first machines were the result of human efforts to improve war-making capabilities; the term engineer
consisting essentially of a solid cylinder, usually of metal, around which an inclined planeinclined plane,
simple machine, consisting of a sloping surface, whose purpose is to reduce the force that must be applied to raise a load. To raise a body vertically a force must be applied that is equal to the weight of the body, i.e.
winds spirally, either clockwise or counterclockwise. It is used to fasten one object to another, to lift a heavy object, or to move an object by a precise amount. The ridge forming the inclined plane is called the thread; in cross section the ridge may be approximately triangular, square, or rounded. The vertical distance from any point on one thread to a corresponding point on the next successive thread is called the pitch. A thread can also be placed on the inner surface of a hollow cylinder. Two screws of the same pitch and diameter, one on the outer surface of a solid cylinder and the other on the inner surface of a hollow cylinder, can be arranged so that one may be driven spirally into the other, as in the common nut and bolt. The thread on the surface of the bolt is called the external, or male, screw; that on the inner surface of the nut, the internal, or female, screw. The common jackscrew used to lift automobiles, houses, and other heavy objects is an application of this principle. The internal screw is situated in the base, the external screw on a metal cylinder; at the top of the cylinder a lever or handle is fastened. As the handle is rotated, the external screw moves up the internal screw and the object placed on top of the jack is lifted. The mechanical advantage of the jackscrew, as of any other screw, is theoretically the ratio between the circumference through which the end of the handle moves and the pitch of the screw. Since, however, there is much friction in the operation of a screw, the amount of workwork,
in physics and mechanics, transfer of energy by a force acting to displace a body. Work is equal to the product of the force and the distance through which it produces movement.
put into this machine is much greater than the amount done and the efficiency is small. On the other hand, the small effort necessary to turn the handle, when compared to the enormous load raised, makes such a device of great value. The screw is often used for making delicate adjustments of tools and machines, e.g., in the micrometer screw and in the carburetor of the gasoline engine (for regulating the flow of gasoline). The self-tapping screw has notches in the first few threads that can cut female threads in a hollow cylinder. Wood and metal screws, the carpenter's and machinist's vise, the propeller of a boat or airplane, Archimedes' screw, and many other devices are applications of the screw.
The following article is from The Great Soviet Encyclopedia (1979). It might be outdated or ideologically biased.
Screw
a machine part of cylindrical—sometimes conical—shape, with a helical surface; or a part with helical blades. Screws are grouped as follows: screws that interact directly with an external, or working, medium, and screws that interact with a threaded opening of another part. The first group includes screws used for the production of rotary motion from moving gases and liquids (for example, in a wind engine); for the production of a tractive force (for example, an airplane propeller or a screw propeller on a ship); and for moving gases, liquids, and viscous, friable, or lump materials, sometimes with the creation in them of pressure and agitation (for example, in fans, pumps, screw conveyors, screw presses, and stirring screw conveyors). The second group includes motion screws, heavy-duty (force) screws, micrometer screws, fastening screws, and set screws.
A motion screw is the basic part of the screw mechanism in lathes and machines used for the rectilinear movement of assemblies and components (supports, tables, and so on) along a set of guides. Motion screws generally have a trapezoidal thread shape—less frequently square, triangular, or special—for screwing on nuts.
A heavy-duty force screw is a main part of a screw mechanism designed to produce large axial stresses in presses, furnace pushers, jacks (load-lifting screws), and so on. Force screws generally have a trapezoidal or buttress thread, less frequently a square thread. Short force screws, such as those in jacks, work by compression; they are checked for strength and buckling during their design and manufacture. Long force screws work by tension and are tested for tensile strength.
Micrometer screws have a precision thread with a small pitch; they are used in measuring machines, devices, and instruments (for example, in micrometers).
A fastening screw is a basic part of a detachable screw joint; the body of the screw has a threaded shaft at one end and a head at the other end. Fastening screws for metals and other solid materials usually have a triangular cylindrical thread. Noncritical small-diameter (to 8 mm) fastening screws are also made as self-tapping screws, with a tapered portion of partially shaped thread on the point. This type of screw extrudes the thread (in soft metals) or cuts it (in plastics and hard metals) when screwed into a smooth hole. Wood screws, which are used for fastening wooden parts, have a tapered portion of the thread at the point. The screw head serves both as a clamp for the parts to be joined and as a grip for tightening the screw with a screwdriver, wrench, or other tool. Standard fastening screws with hexagonal, square, and other types of heads are widespread.
Setscrews are generally used for the precise clamping of geodetic and other types of instruments, as well as for fastening screws in screw joints. Setscrews differ in the shape of the head and the clamping (adjusting) point.
N. IA. NIBERG
What does it mean when you dream about a screw?
The dreamer may feel as if they are being turned like a screw in a situation where someone is taking advantage of them.
screw
[skrü]
(design engineering)
A cylindrical body with a helical groove cut into its surface.
A fastener with continuous ribs on a cylindrical or conical shank and a slotted, recessed, flat, or rounded head. Also known as screw fastener.
McGraw-Hill Dictionary of Scientific & Technical Terms, 6E, Copyright © 2003 by The McGraw-Hill Companies, Inc.
screw
screw: nomenclature
McGraw-Hill Dictionary of Architecture and Construction. Copyright © 2003 by McGraw-Hill Companies, Inc.
screw
1. a device used for fastening materials together, consisting of a threaded and usually tapered shank that has a slotted head by which it may be rotated so as to cut its own thread as it bores through the material
2. a threaded cylindrical rod that engages with a similarly threaded cylindrical hole; bolt
3. a thread in a cylindrical hole corresponding with that on the bolt or screw with which it is designed to engage
4. Billiards snooker
a. a stroke in which the cue ball recoils or moves backward after striking the object ball, made by striking the cue ball below its centre
b. the motion resulting from this stroke
5. another name for propeller
6. Slang an old, unsound, or worthless horse
Collins Discovery Encyclopedia, 1st edition © HarperCollins Publishers 2005
screw
(jargon)
(MIT) A lose, usually in software. Especially used for user-visible misbehaviour caused by a bug or misfeature. This use has become quite widespread outside MIT.
References in periodicals archive ?
Drive the anchor right through the drywall with your hammer and tighten the screw to expand the shank.
"We have to tighten the screws until there's no wiggle room left for terrorists or criminals and no means for them to launch attacks," Avramopoulos told Welt.
ISLAMABAD -- As Middle Eastern countries tighten the screws on human resource exports, Pakistan labour export to Saudi Arabia has plunged by 40 percent in 2017, according to official data available from Bureau of Emigration and Overseas Employment (BOEOE).
The humanitarian crisis in Yemen worsens as the regime in Riyadh and its allies tighten the screws on Yemen's ports.
Haley said the US will "tighten the screws" and putting pressure economically, diplomatically, politically and internationally.
He's incredibly concerned about anything and everything around him, I think this was a message to South Korea after the election and so what we're going to do is continue to tighten the screws. He feels it.
The move comes as the regulator looks to tighten the screws on brokerage houses and also "critically reviews the role of statutory auditors of brokerage houses".
Babe starts to tighten the screws on the Coker family
So, the opportunity to tighten the screws won't come until December.
The law -- Maharashtra Control of Organised Crime Act ( MCOCA) -- will enable the police to tighten the screws on the accused as their confessions before the investigators would now be admissible in court.
New York City Public Advocate, Bill de Blasio, who had teamed up with UANI to launch the Iran Watch List, said, 'Consumers here have the power to force these companies out of Iran and tighten the screws on Tehran's regime.
The ministry pointed out that these practices come as a continuation of the fierce attack against the prisoners after the right-winged extremist Israeli government gave the green light to the prison's administration to increase the pressure on the prisoners and tighten the screws, for political gains.
Site: Follow: Share:
Open / Close
| 2,128
| 9,906
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2020-50
|
latest
|
en
| 0.942309
|
http://v8doc.sas.com/sashtml/ets/chap19/sect27.htm
| 1,516,171,278,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084886830.8/warc/CC-MAIN-20180117063030-20180117083030-00708.warc.gz
| 366,461,318
| 2,583
|
Chapter Contents Previous Next
The SYSLIN Procedure
## TEST Statement
TEST equation , ... , equation / options ;
The TEST statement performs F-tests of linear hypotheses about the parameters in the preceding MODEL statement. Each equation specifies a linear hypothesis to be tested. If more than one equation is specified, the equations are separated by commas.
Variable names must correspond to regressors in the preceding MODEL statement, and each name represents the coefficient of the corresponding regressor. The keyword INTERCEPT is used to refer to the model intercept.
TEST statements can be given labels. The label is used in the printed output to distinguish different tests. Any number of TEST statements can be specified. Labels are specified as follows:
label: TEST ... ;
The following is an example of the use of TEST statement, which tests the hypothesis that the coefficients of X1 and X2 are the same:
proc syslin data=a;
model y = x1 x2;
test x1 = x2;
run;
The following statements perform F-tests for the hypothesis that the coefficients of X1 and X2 are equal, and that the sum of the X1 and X2 coefficients is twice the intercept, and for the joint hypothesis.
proc syslin data=a;
model y = x1 x2;
x1eqx2: test x1 = x2;
sumeq2i: test x1 + x2 = 2 * intercept;
joint: test x1 = x2, x1 + x2 = 2 * intercept;
run;
The following are additional examples of TEST statements:
test x1 + x2 = 1;
test x1 = x2 = x3 = 1;
test 2 * x1 = x2 + x3, intercept + x4 = 0;
test 2 * x1 - x2;
The TEST statement performs an F-test for the joint hypotheses specified. The hypothesis is represented in matrix notation as follows:
The F test is computed as
where b is the estimate of , m is the number of restrictions, and is the model mean square error. See the section "Computational Details" for information on the matrix X'X.
The test performed is exact only for ordinary least squares, given the OLS assumptions of the linear model. For other estimation methods, the F-test is based on large sample theory and is only approximate in finite samples.
If RESTRICT or SRESTRICT statements are used, the tests computed by the TEST statement are conditional on the restrictions specified. The validity of the tests may be compromised if incorrect restrictions are imposed on the estimates.
The PRINT option can be specified in the TEST statement after a slash (/):
PRINT
prints intermediate calculations for the hypothesis tests.
Note: The TEST statement is not supported for the FIML estimation method.
Chapter Contents Previous Next Top
| 573
| 2,563
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.75
| 3
|
CC-MAIN-2018-05
|
latest
|
en
| 0.896017
|
https://byjus.com/cbse/class-10-maths-objective-questions-chapter-9-applications-of-trigonometry/
| 1,696,229,522,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233510983.45/warc/CC-MAIN-20231002064957-20231002094957-00620.warc.gz
| 177,955,197
| 120,760
|
# CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions
CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions deal with how trigonometry helps to find the height and distance of different objects without any measurement. In earlier days, astronomers used trigonometry to calculate the distance from the earth to the planets and stars. Trigonometry is also mostly used in navigation and geography to locate the position in relation to the latitude and longitude. Keeping in mind the latest modification of the CBSE exam pattern, we have compiled the CBSE Class 10 Maths Chapter 9 Applications of Trigonometry Objective Questions for the students to prepare well for the board exams. Students will learn the applications of trigonometry with real-life examples in a better way with the help of the MCQs provided in this article.
Solving these CBSE Class 10 Maths Chapter 9 objective questions is a good practice as it provides additional questions for the students to practise for the board exams. Students also get acquainted with a wide range of questions of different types and difficulty levels by practising these objective questions.
Find below the CBSE Class 10 Maths Objective Questions and the sub-topics it covers.
### List of Sub-topics Covered in Chapter 9
We have compiled here some 20 multiple-choice questions covering the below-given topics from the chapter.
9.1 Applications of Trigonometry (6 MCQs from the Topic)
9.2 Introduction (4 MCQs from the Topic)
9.3 Heights and Distances (10 MCQs from the Topic)
## Download Free CBSE Class 10 Maths Chapter 9 – Applications of Trigonometry Objective Questions PDF
### Applications of Trigonometry
1. A technician has to repair a light on a pole of a height of 10 m. She needs to reach a point 1 m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use when inclined at an angle of 60∘ to the ground that would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder?
Solution: The given situation is explained in the figure below:
DB – DC = CB
⇒BC=9msin60°=BCAC⇒AC=BCsin60°=
∴ The height of the ladder should be 6√3 m.
1. A statue, 2 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60°, and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Solution:
CD = 2m
Let BC = x
AB = x (using tan 45∘)
In ΔABD,
BD = AB √3 = x √3
We know that CD = BD – BC = 2m
X(√3 – 1)
X = 2(√3 -1 )
Therefore, the height of the pedestal is 2(√3 -1 ).
1. The angle of elevation of the top of a building from the foot of the tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 60 m high, find the height of the building.
1. 30 m
2. 40 m
3. 20 m
4. 10 m
Solution:
The given situation is explained in the figure above.
tan60°=DC/BC⇒BC=DC/tan60°= 60/ √3 = 20 √3 m
tan30°=AB/BC
⇒AB=BC×tan30°=20 √3 × (1/√3 m) = 20 m
Thus, the height of the building is 20 m.
1. A TV tower stands vertically on a bank of a canal, with a height of 10 √3 m. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the distance between the opposite bank of the canal at the point with a 30° angle of elevation.
1. 30 m
2. 20 m
3. 45 m
4. 35 m
Solution:
The above figure represents the situation given in the question.
tan60°=AB/ BC
⇒AB=BCtan60°= BC √3……….(1)
⇒BC=AB/tan60°=AB/ tan60°
tan30°=AB/BD=AB/ (CD+BC)
⇒DC+BC=AB/tan30°=AB √3
⇒DC=AB
⇒DC = 20 m
Hence, the distance between the opposite bank of the canal is 20 m.
1. As observed from the top of a 150 m high lighthouse from the sea level, the angles of depression of the two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Answer: (B) 150 (√3 – 1)
Solution:
Here, the lighthouse, BC = 150 m
In ΔBDC,
BD = BC = 150 m (using tan 45°)
In ΔABC,
AB = BC √3 (using tan 30°)
AB= 150√3
So, distance AD = 150 (√3 – 1)
Hence, the between the two ships is 150 (√3 – 1).
1. An observer of √3m tall is 3 m away from a pole of 2√3 high. What is the angle of elevation of the top of the pole from the observer?
1. 60°
2. 30°
3. 45°
4. 90°
Solution:
Height of the pole that is above man’s height = 2 √3 – √3 = √3 m
Hence, AB = √3 m
BC = 3 m
Hence, tan C = AB/BC =
⟹ C, angle of elevation = 30°
Therefore, the angle of elevation of the top of the pole from the observer is 30°.
### Introduction
1. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution: In ΔABC, taking the tangent of ∠C, we have,
tan C=AB/BC
⇒tan 30°=h/30
⇒
$$\begin{array}{l}\frac{1}{\sqrt{3}}\end{array}$$
= h/30
=10(1.732)
=17.32m = 10√3 m
Hence, the height of the tower is 10√3 metres.
1. An observer having 1.5 m tall is 28.5 m away from a tower. The angle of elevation of the top of the tower from her eyes is 45°. What is the height of the tower?
1. 20 m
2. 10 m
3. 40 m
4. 30 m
Solution: Let AB be the tower of height h, and CD be the observer of height, 1.5 m at a distance of 28.5 m from tower AB.
In ΔAED, we have
tan45°=h/28.5
⇒1=h/28.5
⇒h=28.5 m
∴ h=AE+BE=AE+DC
= (28.5+1.5) m=30Â m
Height of tower = h + 1.5
= 28.5 + 1.5
= 30 m
Hence, the height of the tower is 30 m.
1. An electrician has to repair an electrical fault on a pole of a height of 4 m. He needs to reach a point 1.3 m below the top of the pole to undertake the repair work. What should be the length of the ladder that he should use when inclined at an angle of 60° to the horizontal that would enable him to reach the required position?
Solution: Let AC be the electric pole of height 4 m. Let B be a point 1.3 m below the top A of the pole AC.
Then, BC = AC – AB = (4 – 1.3) m = 2.7 m
Let BD be the ladder inclined at an angle of 60° to the horizontal.
In ΔBCD, we have
sin60°=2.7/L.
Hence, the length of the ladder should be m.
1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 30°.
1. 10 m
2. 15 m
3. 20 m
4. 35 m
Solution: Let AB be the vertical pole and CA be the 20 m long rope, such that one end is tied from the top of the vertical pole AB and the other end C is tied to a point C on the ground.
In ΔABC, we have
sin 30°=h/20
⇒1/2=h/20
⇒h=10 m
Hence, the height of the pole is 10 m.
### Heights and Distances
1. An observer of 2.25 m tall is 42.75 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
1. 40 m
2. 50 m
3. 45 m
4. 35 m
Solution:
The given situation is explained in the figure above:
In triangle ABE,
tan45°=AB/EB
Also, EB=DC
∴ tan45°=AB/ DC
⇒AB=DC × tan 45°
⇒AB=1×42.75
Hence, the height of the chimney =Â ACÂ = AB + BC
We can observe that BC = ED.
Thus, AC = AB + ED
= 42.75 + 2.25
= 45 m.
Hence, the height of the chimney is 45 m.
1. A tower stands vertically on the ground. From a point on the ground, which is 30 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 30°. Find the height of the tower.
Solution: The given situation is explained in the Δ below:
Now, tan30°=AB/ BC
∴ The height of the tower is 10√3 m.
1. The angles of depression of the top and the bottom of a 10 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi-storeyed building.
Solution:
The above figure represents the situation aptly.
(ED is the height of the building)
1. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at the height of 2 m and is inclined at an angle of 30° to the ground. What should be the length of the slide?
1. 4
2. 2
3. 1.5
4. 3
Solution: The given situation can be explained using the figure below:
In the right-angled triangle ABC,
sin∠ABC=AC/AB=1/2
⇒sin30°=2/AB⇒AB= 2/ (1/2)
⇒AB = 4m
∴ The length of the slide is 4m.
1. A kite is flying at a height of 30 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Solution: The situation can be explained using the figure below:
In the given right-angled triangle:
sin (∠ACB) =AB/AC
⇒sin60°=AB/AC
⇒AC=AB/sin60°=
∴ The length of the string is 20√3 m
1. A vertical pole of 30 m is fixed on a tower. From a point on the level ground, the angles of elevation of the top and bottom of the pole are 60° and 45°. Find the height of the tower.
Answer: (B) 15 (√3 + 1)
Solution:
The situation can be explained by the figure above.
Therefore, the height of the tower is 15 (√3 + 1).
1. The value of tan A +sin A=M, and tan A – sin A=N.
The value of (M2−N2) / (MN) 0.5
1. 4
2. 3
3. 2
4. 1
Solution: M2-N2 = (Tan A+ Sin A + Tan A –Sin A) (Tan A +Sin A – Tan A+ Sin A)
M2-N2Â =4 tan A sin A
And (MN) 0.5 = (tan2A−sin2A) 0.5
1. Two towers, A and B, are standing at some distance apart. From the top of tower A, the angle of depression of the foot of tower B is found to be 30°. From the top of tower B, the angle of depression of the foot of tower A is found to be 60°. If the height of tower B is ‘h’ m, then the height of tower A in terms of ‘h’ is _____ m
Solution:
Let the height of tower A be = AB =Â H.
And the height of tower B = CD = h
In triangle ABC,
tan30° = AB/AC = H/AC ……………………………. 1
tan60° = CD/AC = h/AC………………………………….2
Divide 1 by 2
We get tan30°/tan60° = H/h
Therefore, H = h/3 m.
1. A 1.5 m tall boy is standing some distance from a 31.5 m tall building. If he walks ’d’ m towards the building, the angle of elevation of the top of the building changes from 30° to 60°. Find the length of d. (Take √3 = 1.73)
1. 30.15 m
2. 38.33 m
3. 22.91 m
4. 34.55 m
Solution:
The above figure explains the situation given in the question.
∴ The distance moved by the boy, d, is 34.55 m.
1. The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45° and 30°. Find the distance between the two objects. (Take √3 = 1.73,)
1. 200 m
2. 150 m
3. 107.5 m
4. 73.2 m
Solution: Let C and D be the objects and CD be the distance between the objects.
In ΔABC, tan 45° = AB/AC
AB=AC=100Â m
In ΔABD, tan 30° = AB/AD
Hence, the distance between the two objects is 73.2 metres.
You can access the PDF link to download multiple-choice questions from Chapter 9 – Applications of Trigonometry, categorised as per the topics from which it is taken. Solving these questions will help the students to score better in the board exams because, as per the latest exam pattern, more objective questions are expected to be included in question papers.
### CBSE Class 10 Maths Chapter 9 Extra MCQs
1. If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, calculate the angle of elevation of the cloud.Â
(a) equal to the angle of depression of its reflection.
(b) double to the angle of depression of its reflection
(c) not equal to the angle of depression of its reflection
(d) information insufficient
2. There are two windows in the house. A window of the house is at a height of 1.5 m above the ground, and the other window is 3 m vertically above the lower window. Ram and Shyam are sitting inside the two windows. At an instant, the angle of elevation of a balloon from these windows is observed as 45° and 30°, respectively. Then, what is the height of the balloon from the ground?
(a)Â 7.269 m
(b) 8.269 m
(c)Â 7.598 m
(d) 8.598 m
Apart from these MCQs, students can also download other resources at BYJU’S like NCERT Solutions, previous years’ papers, Syllabus, study notes, sample papers, and important questions for all the classes to prepare more efficiently for the exams. Keep learning and stay tuned for further updates on CBSE.
| 4,118
| 12,896
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.6875
| 5
|
CC-MAIN-2023-40
|
latest
|
en
| 0.905475
|
http://mathhelpforum.com/algebra/218451-prove-srt-ab-b-2-a.html
| 1,527,144,789,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865928.45/warc/CC-MAIN-20180524053902-20180524073902-00071.warc.gz
| 196,408,192
| 12,310
|
# Thread: prove srt(ab) => (a+b)/2
1. ## prove srt(ab) => (a+b)/2
how would I go about proving this?
$\displaystyle sqrt(ab) \geq (a+b)/2$
i manipulated the expression so that $\displaystyle 0 \geq (a-b)^2$ but im not sure if that proves anything
2. ## Re: prove srt(ab) => (a+b)/2
Originally Posted by kingsolomonsgrave
how would I go about proving this?
$\displaystyle sqrt(ab) \geq (a+b)/2$
i manipulated the expression so that $\displaystyle 0 \geq (a-b)^2$ but im not sure if that proves anything
You have it written incorrectly.
$\displaystyle \\(x-y)^2=x^2-2xy+y^2\ge 0\\x^2+y^2\ge 2xy$
If $\displaystyle a\ge 0~\&~b\ge 0$ then let $\displaystyle x=\sqrt a~\&~b=\sqrt b$.
3. ## Re: prove srt(ab) => (a+b)/2
Originally Posted by kingsolomonsgrave
how would I go about proving this?
$\displaystyle sqrt(ab) \geq (a+b)/2$
i manipulated the expression so that $\displaystyle 0 \geq (a-b)^2$ but im not sure if that proves anything
Are you sure your original equation wasn't \displaystyle \displaystyle \begin{align*} \sqrt{ab} \leq \frac{a + b}{2} \end{align*}?
4. ## Re: prove srt(ab) => (a+b)/2
yes! I knew something was off. It should have been:
it should have been:
Prove $\displaystyle \sqrt{ab}\leq \frac{a+b}{2}$ for all positive integers.
5. ## Re: prove srt(ab) => (a+b)/2
Originally Posted by kingsolomonsgrave
yes! I knew something was off. It should have been:
it should have been:
Prove $\displaystyle \sqrt{ab}\leq \frac{a+b}{2}$ for all positive integers.
\displaystyle \displaystyle \begin{align*} 0 &\leq \left( a - b \right) ^2 \\ 0 &\leq a^2 - 2ab + b^2 \\ 4ab &\leq a^2 + 2ab + b^2 \\ 4ab &\leq \left( a + b \right) ^2 \\ ab &\leq \frac{ \left( a + b \right) ^2}{4} \\ ab &\leq \left( \frac{a + b}{2} \right) ^2 \\ \sqrt{ab} &\leq \frac{a + b}{2} \end{align*}
6. ## Re: prove srt(ab) => (a+b)/2
geometrically
7. ## Re: prove srt(ab) => (a+b)/2
How do we know the vertical line is sqrt(ab)?
8. ## Re: prove srt(ab) => (a+b)/2
Originally Posted by kingsolomonsgrave
How do we know the vertical line is sqrt(ab)?
That's because the two small right triangles into which the big triangle is split are similar.
9. ## Re: prove srt(ab) => (a+b)/2
Thanks! How does that tell me that that length is $\displaystyle \sqrt{ab}$?
I can see that the smaller triangle is proportional to the larger and that they have the same angles.
The most I can come up with is that if a = b then you get a big right triangle with angles of 45 on either side. That way $\displaystyle a^2+a^2= \sqrt{2}$ and $\displaystyle b^2+b^2 = \sqrt{2}$ and $\displaystyle a=b so a^2=ab=b^2$ so the line is $\displaystyle \sqrt{ab}$
10. ## Re: prove srt(ab) => (a+b)/2
Originally Posted by kingsolomonsgrave
How does that tell me that that length is $\displaystyle \sqrt{ab}$?
If the altitude of the big triangle is h, then it follows from the similarity of small triangles that a / h = h / b.
11. ## Re: prove srt(ab) => (a+b)/2
Originally Posted by kingsolomonsgrave
Thanks! How does that tell me that that length is $\displaystyle \sqrt{ab}$?
Have you heard the term mean proportional ?
If $\displaystyle \frac{a}{h}=\frac{h}{b}>0$ then $\displaystyle h$ is the mean proportional between $\displaystyle a~\&~b$.
In a right triangle the altitude from the right angle to the hypotenuse is the mean proportional between the parts.
That true in the supplied diagram .
| 1,097
| 3,385
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.15625
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.699295
|
http://www.businessrefinery.com/b/21/85/
| 1,371,640,307,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368708711794/warc/CC-MAIN-20130516125151-00079-ip-10-60-113-184.ec2.internal.warc.gz
| 349,592,962
| 5,433
|
# Part III: Examining Changes in Terms of Energy in VS .NET Making European Article Number 13 in VS .NET Part III: Examining Changes in Terms of Energy
Part III: Examining Changes in Terms of Energy
Encoding EAN13 In Visual Studio .NET
Using Barcode maker for .NET Control to generate, create EAN 13 image in .NET framework applications.
Consider the following reaction: N2(g) + O2(g) 2NO(g) At equilibrium, you measure the following partial pressures: P(N2) = 476 10 2 atm, P(O2) = 982 10 3 atm, and P(NO) = 263 10 7 atm a What is the Keq for this reaction b If you measured P(O2) = 374 10 2 atm, and other partial pressures remained unchanged, what is the reaction quotient for the reaction In which direction would the reaction proceed
Reading EAN / UCC - 13 In VS .NET
Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET framework applications.
You perform the following reaction: 2A + 2B 3C + D After waiting three hours, you measure the following concentrations: [A] = 273 mM, [B] = 347 mM, [C] = 0443M, and [D] = 789 mM a What is the reaction quotient for the system b If the Keq for this reaction is 185 102, has the reaction completed
Bar Code Creator In Visual Studio .NET
Using Barcode maker for .NET framework Control to generate, create barcode image in .NET framework applications.
Solve It
Bar Code Decoder In .NET
Using Barcode decoder for .NET framework Control to read, scan read, scan image in Visual Studio .NET applications.
Solve It
EAN 13 Creation In C#.NET
Using Barcode printer for .NET framework Control to generate, create EAN-13 image in .NET applications.
Consider the reaction A + B 2C The free energy change for this reaction is 258 kJ mol 1, and the enthalpy change for the same reaction is 123 kJ mol 1 a At 273K, what is the entropy change for the reaction What drives the reaction forward favorable enthalpy change, favorable entropy change, or both b What is the expression for Keq for the reaction, and what is its value c If a reaction mixture at equilibrium contains 0744M reactant A and 117M product C, what is the concentration of reactant B
EAN13 Maker In VB.NET
Using Barcode encoder for .NET Control to generate, create EAN / UCC - 13 image in VS .NET applications.
Solve It
Creating Bar Code In .NET
Using Barcode printer for VS .NET Control to generate, create barcode image in .NET applications.
14: Exploring Rate and Equilibrium
Printing Code-39 In VS .NET
Using Barcode printer for .NET framework Control to generate, create Code-39 image in .NET framework applications.
Checking Out Factors that Disrupt Equilibrium
Painting Bar Code In Visual Studio .NET
Using Barcode generation for .NET Control to generate, create bar code image in .NET framework applications.
After a chemical system has reached equilibrium, that equilibrium can be disrupted, or perturbed Think of systems at equilibrium as people who have finally found their easy chair at the end of a long day You may rouse them to take out the trash, but they ll return to the easy chair at the first opportunity This concept is more or less the idea behind Le Chatelier s Principle: The equilibrium of a perturbed system shifts in the direction that minimizes the perturbation Perturbations include changes in concentration, pressure, and temperature Concentration: If a system is at equilibrium, Q = Keq Adding or removing reactant or product disrupts the equilibrium such that Q < Keq or Q > Keq The equilibrium reasserts itself in response If reactant is added or product is removed, reactant converts into product If product is added or reactant is removed, product converts into reactant Either way, chemistry occurs until Q = Keq once more The equilibrium shifts to oppose the perturbation Pressure: Reactions that include gases as reactants and/or products are particularly sensitive to pressure perturbation If pressure is suddenly increased, equilibrium shifts toward the side of the reaction that contains fewer moles of gas, thereby decreasing pressure If pressure is suddenly decreased, equilibrium shifts toward the side of the reaction that contains more moles of gas, thereby increasing pressure Consider the following reaction: N2(g) + 3H2(g) 2NH3(g) A given amount of mass on the reactant side of the equation (as N2 and H2) corresponds to double the moles of gas as the same mass on the product side (as NH3) Imagine that the system is at equilibrium at a low pressure Now imagine that the pressure suddenly increases, perturbing that equilibrium Reactant (N2 and H2) converts to product (NH3) so the total moles of gas decrease, thereby lowering the pressure If the system suddenly shifts to lower pressure, NH3 converts to N2 and H2, so the total moles of gas increase, thereby raising the pressure The equilibrium shifts to oppose the perturbation Temperature: Reactions that absorb or give off heat (that is, most reactions) can be perturbed from equilibrium by changes in temperature The easiest way to understands this behavior is to explicitly include heat as a reactant or product in the reaction equation: A + B C + heat Imagine that this reaction is at equilibrium Now imagine that the temperature suddenly increases Product C absorbs heat, converting to reactants A and B Because the heat product has been decreased, the temperature of the system decreases If the temperature suddenly shifts down from equilibrium, reactants A and B convert to product C, releasing heat The released heat increases the temperature of the system The equilibrium shifts to oppose the perturbation
Bar Code Maker In VS .NET
Using Barcode printer for VS .NET Control to generate, create barcode image in Visual Studio .NET applications.
Encoding Leitcode In .NET
Using Barcode printer for .NET Control to generate, create Leitcode image in Visual Studio .NET applications.
UPC A Scanner In .NET Framework
Using Barcode scanner for .NET framework Control to read, scan read, scan image in .NET framework applications.
Creating UCC - 12 In Java
Using Barcode generation for Java Control to generate, create UCC.EAN - 128 image in Java applications.
UCC.EAN - 128 Generator In C#.NET
Using Barcode generator for Visual Studio .NET Control to generate, create GTIN - 128 image in VS .NET applications.
Encoding USS Code 39 In VB.NET
Using Barcode encoder for Visual Studio .NET Control to generate, create Code 39 image in VS .NET applications.
| 1,438
| 6,397
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.34375
| 3
|
CC-MAIN-2013-20
|
latest
|
en
| 0.846071
|
http://mathhelpforum.com/algebra/167610-algebra-problems-print.html
| 1,527,077,316,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00434.warc.gz
| 186,349,013
| 3,758
|
# Algebra problems
• Jan 6th 2011, 09:45 AM
eagertolearn
Algebra problems
I have 3 problems I need help with. I am completely clueless but more than willing to learn. I also don't know how to insert some of the symbols so I wrote it out, I hope it's not to confusing
1. x(not equal to sign)0, then 1-1
x
2. a(not equal to) 0, a(not equal to)1; ax-b then b=
a-b
3. k= 1 mb(squared), then b=
2
Thank You !!
Edit: For some reason when I submitted the post some of the numbers that were supposed to go underneath the underline moved to the left.
1. x is supposed to go under 1-1
2. a-b is supposed to go under ax-b
3. 2 is supposed to under the 1 as to say half
• Jan 6th 2011, 09:55 AM
e^(i*pi)
1. $\displaystyle \dfrac{1-1}{x} \ ,\ x \neq 0$ Unless I'm missing something very obvious this is 0.
2. $\displaystyle \dfrac{ax-b}{a-b}$ it needs to be equal to something to manipulate it. I will say it's equal to y for argument's sake
$\displaystyle y = \dfrac{ax-b}{a-b}$ then $\displaystyle y(a-b) = ax-b$ and $\displaystyle ya-yb = ax-b$.
Collect the b terms on one side and factor.
3. $\displaystyle \frac{1}{2}mb^2$. Again it needs to equal to something.
Once more assuming it's equal to y then multiplying through by 2/m gives $\displaystyle \dfrac{2y}{m} = b^2$
• Jan 6th 2011, 09:59 AM
snowtea
Your post is really hard to read (perhaps a lot of spaces were ignored).
You may want to read the latex tutorial in the Latex Help subforum for future posts.
As an example type:
[tex] x \neq \frac{y^2 + 1}{z + a} [/ math]
(without space after the /)
to get
$\displaystyle x \neq \frac{y^2 + 1}{z + a}$
• Jan 6th 2011, 01:23 PM
eagertolearn
Quote:
Originally Posted by e^(i*pi)
1. $\displaystyle \dfrac{1-1}{x} \ ,\ x \neq 0$ Unless I'm missing something very obvious this is 0.
2. $\displaystyle \dfrac{ax-b}{a-b}$ it needs to be equal to something to manipulate it. I will say it's equal to y for argument's sake
$\displaystyle y = \dfrac{ax-b}{a-b}$ then $\displaystyle y(a-b) = ax-b$ and $\displaystyle ya-yb = ax-b$.
Collect the b terms on one side and factor.
3. $\displaystyle \frac{1}{2}mb^2$. Again it needs to equal to something.
Once more assuming it's equal to y then multiplying through by 2/m gives $\displaystyle \dfrac{2y}{m} = b^2$
Thank you so much for your help! I also said the same thing for #1 that it IS 0 but the problem as was given to me clearly states x is not equal to 0. So clearly I was confused.
Also for #2 and #3 in order to find b which is what the original problems are asking me to do, you basically are isolating and getting b by itself correct?
Again thanks so much.
• Jan 6th 2011, 01:25 PM
eagertolearn
Quote:
Originally Posted by snowtea
Your post is really hard to read (perhaps a lot of spaces were ignored).
You may want to read the latex tutorial in the Latex Help subforum for future posts.
As an example type:
[tex] x \neq \frac{y^2 + 1}{z + a} [/ math]
(without space after the /)
to get
$\displaystyle x \neq \frac{y^2 + 1}{z + a}$
Thanks!
• Jan 6th 2011, 01:42 PM
e^(i*pi)
Quote:
Originally Posted by eagertolearn
Thank you so much for your help! I also said the same thing for #1 that it IS 0 but the problem as was given to me clearly states x is not equal to 0. So clearly I was confused.
Also for #2 and #3 in order to find b which is what the original problems are asking me to do, you basically are isolating and getting b by itself correct?
Again thanks so much.
For 2 and 3 you have an expression rather than an equation because you've not told us what the formulae are equal to. You can't get b on it's own because there is nothing on the other side of the equals sign to manipulate
For formatting, if you can't use latex then at least use plain text:
(ax-b)/(a-b) is question 2 for example
| 1,148
| 3,775
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2018-22
|
latest
|
en
| 0.887021
|
http://nzmaths.co.nz/resource/adding-ten-tiles-ii
| 1,369,066,916,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2013-20/segments/1368699068791/warc/CC-MAIN-20130516101108-00036-ip-10-60-113-184.ec2.internal.warc.gz
| 184,122,564
| 13,944
|
Te Kete Ipurangi
Communities
Schools
### Te Kete Ipurangi user options:
Level Six > Number and Algebra
Achievement Objectives:
Specific Learning Outcomes:
Use the numbers facts with digits to solve a problem
Devise and use problem solving strategies to explore situations mathematically (guess and check, be systematic, look for patterns, think, use algebra).
Description of mathematics:
This problem needs a certain amount of experimentation to get it out. In the process students will be practising their basic number facts. However, to get the complete answer required by the Extension probably needs a little algebra.
This problem links strongly to both Adding Ten Tiles I and the Nine Tiles problem. We suggest that your students try Adding Ten Tiles first as there are fewer number to play with. The method that worked for that problem will also work for this one.
Required Resource Materials:
Copymaster of the problem (English)
Copymaster of the problem (Māori)
Activity:
### The Problem
Jim still has ten tiles with one of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 on each of them. He plays around and discovers that he can make some addition sums like the one below. The four digits in the answer to his sums always seem to add up to 18.
How many answers like this can Jim make?
### Teaching sequence
1. Start with the class in their groups and either pose the problem to the whole class or give each group a copy of the problem and tell them to begin. Check that they all understand that there are ten different numbers (0 to 9 inclusive) that have to be used in Jim’s problem.
2. Allow the students to guess at first and record their answers on the board as they find them. For groups who seem to be stuck, ask
Where do you think the number 1 goes? Why?
What four numbers add up to 18? So what could just the answer to Jim’s sum be?
3. For groups that have managed to guess a few answers ask
How did you get the answer?
4. When two different addition answers with the same digits in the sum have been found ask
What are the similarities between these answers?
Can you get another addition sum using the same numbers? How many?
5. When a number of answers have been found, get the class’s attention. Ask
Have we got all of the possible answers?
How will we know when we have all of the answers?
6. When the class has realised that there are a number of ways of getting 18, split these ways amongst the different groups. This will speed up the process of finding the overall answers.
7. Ask some of the groups to explain what they have done. Get all groups to write up the problem as far as they were able to go, giving reasons for what they have done.
#### Extension to the problem
Jane is sick of pussyfooting around. She has been working on Jim’s problem in Adding Ten Tiles. She wants to know if there are any sums that have four-digit answers that don’t add up to 9 or 18. Can you help her?
### Solution
To start this problem off we would be inclined to guess and check and hope that something turns up. The full set of answers is:
3 2 4 3 4 2 4 3 2 4 2 3 7 6 5 7 5 6 6 5 7 6 7 5 1 0 8 9 1 0 9 8 1 0 8 9 1 0 9 8 (i) (ii) (iii) (iv)
We go through a complete solution for Jim’s problem.
Step 1: The first thing to notice is the value of the left-most digit in the answer. It couldn’t be 0 as then the answer wouldn’t be a four-digit answer. In that case we’d be looking at the problem Nine Tiles.
Now to get that digit we have to add two other digits plus a possible carry-over from the tens column. The biggest two digits we could have are 8 and 9. The carry-over cannot be bigger than 2. The sum of 8 and 9 and 2 is 19, which is less than 20. So the carry-over from the hundreds column is 1. This is the value of the left-most number in the answer.
Step 2: So now we have to decide what three digits plus 1 add up to 18. Working systematically we get: 9, 8, 1, 0; 9, 6, 2, 1; 9, 5, 3, 1; 8, 7, 2, 1; 8, 6, 3, 1; 7, 6, 4, 1. (Other possibilities don’t include 1 or contain repeated digits.)
Step 3: 9, 8, 1, 0:To get the answers to this question we need to go through all of the various answer digit combinations. We’ll start with 9, 8, 1, 0. Remember that 1 is the left-most digit. Since 9 and 8 are used in the answer to the sum they can’t be used for the addends (the numbers that are being added together). So 9 and 8 can’t be in the hundreds position.
So we are looking at 1089 or 1098. We’ll do 1089 in full and leave1098 to you.
First notice that there are no carry-overs in the units or tens columns as we do not have big enough digits left to add to 19 or 18, respectively. So we’ll concentrate on the hundreds column. How can we get 10? The only two possibilities are 3 and 7 and 4 and 6. (Any other possibilities include digits that have been used already or repeated digits.)
Let’s go with 3 and 7. To get a 9 in the units position we need to add 2 and 7, 3 and 6 or 4 and 5. But the only legal one of these is 4 and 5. At this stage the only two digits left are 2 and 6. Fortunately they give us the 8 that we need. So the Jim’s answer that we get here is (i) or some rearrangement of the numbers in the columns making up the addends.
Now if we try 4 and 6, we have to use 2 and 7 in the units column and 3 and 5 in the tens column. This gives us (iii) or some slight rearrangement of it.
As we said, we’ll leave 1098 to you but you should get (ii) and (iv) using the method above.
Step 4: 9, 6, 2, 1: As above 9 can’t be the hundreds digit of the answer. So we have to have 1269, 1296, 1629 or 1692.
Consider 1269. We can get 9 by adding 0 + 9, 1 + 8, 2 + 7, 3 + 6 or 4 + 5. The last one is the only legal sum. To get 6 we need 0 + 6, 1 + 5 or 2 + 4. But none of these is legal and so 1269 can’t be one of Jim’s answers.
Consider 1296. To get 6 we would need 0 + 6, 1 + 5 or 2 + 4. None of these is legal.
Consider 1629. The only legal way to get 9 is 4 + 5. To get 2 we need 0 + 2 or 1 + 1 but neither is legal.
Consider 1692. To get 2 we need 0 + 2 or 1 + 1 but neither is legal.
Step 5: 9, 5, 3, 1: Once again we can only have 1359, 1395, 1539 or 1593 as we can’t have 9 in the hundreds position.
Consider 1359 and 1395. In both cases there are no carry-overs for the 5. To get 5 we need 0 + 5, 1 + 4 or 2 + 3, none of which are legal.
A similar argument applies to the 3 in 1539 and 1593.
Step 6: 8, 7, 2, 1: Arguing as we did before on the 9, 8 can’t be in the hundreds column and neither can 7. So we are left with 1278 or 1287. Now 9 can’t be in the units or hundreds column of an addend in either case as there is no digit left that will take it up to 17 or 18.
To make 7 we need 0 + 7, 1 + 6, 2 + 5 or 3 + 4 and to make 8 we need 0 + 8, 1 + 7, 2 + 6 or 3 + 5. For 7, 3 + 4 is the only legal possibility while for 8 it’s 3 + 5. This would require the use to two 3s.
Step 7: 8, 6, 3, 1: Again 8 can’t be in the hundreds position.
Consider 1638 and 1683. To get 6 in the hundreds position we’d need 9 + 7 and no carry-over (9 + 6 and 8 + 7 are illegal). But 3 has to have a carry over since 0 + 3 and 1 + 2 are illegal. That rules out 1638. To get 13 we need 6 + 7, 5 + 8 or 4 + 9. None of these is legal.
Step 8: 7, 6, 4, 1: As above 7 can’t be the hundreds digit. To get 6 there we’d need 9 + 7 but that isn’t possible. So we reduce the possibilities to 1467 or 1476. By what we have just said, neither the 6 nor the 7 can be associated with a carry-over. So 6 = 2 + 4, while 7 = 2 + 5 or 3 + 4. These conditions are incompatible so tis case is complete.
Step 9: In summary we only have the four answers to Jim’s problem listed at the start of this solution. To remind you they are
3 2 4 3 4 2 4 3 2 4 2 3 7 6 5 7 5 6 6 5 7 6 7 5 1 0 8 9 1 0 9 8 1 0 8 9 1 0 9 8 (i) (ii) (iii) (iv)
Of course there is also the possibility of moving digits vertically up and down above the horizontal line to give 32 answers altogether.
#### Solution to the extension
Now there are clearly some patterns in the above answers that ought to be exploitable but we can’t see how. Hence we’ll go the algebra route and see what happens.
First note that the left-hand digit of the bottom row has to be 1 as the numbers are not big enough for it to be any larger. (We also ignore the possibility of this number being 0 as that would take us back to the Nine Tiles problem.)
Suppose that we have an answer of the form
a b c d e f 1 h i j
Now c + f = j or 10 + j, b + e or b + e + 1 = i or 10 + i, and a + d or a + d + 1 = 10 + h. Assuming that there are r carry-overs, where r = 1, 2 or 3, we can add these equations up to get
a + b + c + d + e + f + (r – 1) = 10r + h + i + j
[45 – (1 + h + i + j)] + (r – 1) = 10r + h + i + j
44 + (r – 1) – 10r = 2(h + i + j)
This means that r has to be odd, so r = 1 or 3.
If r = 1: Then 44 – 10 = 2(h + i + j). So h + i + j = 17.
Now where can the zero be? It can’t be b, e, c or f as that would mean that we would need to repeat one of these numbers in the answer. Suppose that it is a. Then a + d has to carry over. But this is not possible as 0 + d can never be bigger than 9 (remember there is no carry over from the tens column). So one of h, i or j has to be zero. The only way that three numbers can add to 17, where one is zero, is for the numbers to be 9, 8 and 0.
The zero has to be h as a + d can never be 18 or 19 (without a carry-over). This means that the number in the bottom line has to be 1089 or 1098. So we have to get 8 and 9 without carry-overs and two numbers have to add to 10. We put down all the relevant sums that don’t use 0, 1, 8 or 9.
8 = 2 + 6 = 3 + 5; 9 = 2 + 7 = 3 + 6 = 4 + 5; 10 = 3 + 7 = 4 + 6.
If 8 = 2 + 6, then 9 = 4 + 5 and 10 = 3 + 7. This gives us the answers (i) and (ii).
If 8 = 3 + 5, then 9 = 2 + 7 and 10 = 4 + 6. This gives us the answers (iii) and (iv).
(i) 324 756 1089 (ii) 324 756 1098 (iii) 432 657 1089 (iv) 423 675 1098 (v) 437 589 1026 (vi) 347 859 1206 (vii) 473 589 1062 (viii) 743 859 1602 (ix) 246 589 1035 (x) 426 879 1305 (xi) 264 789 1053 (xii) 624 879 1503
If r = 3: Then 16 = 2(h + i + j). So h + i + j = 8.
Now 8 can be made using three digits, none of which is 1, in only the following two ways:
8 = 0 + 2 + 6 = 0 + 3 + 5.
So the number in the bottom line has to be 1026 (or 1 and some permutation of the last three digits) or 1035 (or 1 and some permutation of the last three digits).
Let’s try 1026. Then we need to get 16 by adding two digits. This can be done using 9 + 7 only. So c, f are 7, 9 in any order. To get 12 in the tens column we have the carry-over so we must get 11. This can be done using 6 + 5 or 7 + 4 or 8 + 3 or 9 + 2. But each of those pairs has had one number used already except 8 + 3, so b, e are 3, 8, in some order. Finally to get 10 including the carry-over we have to have two digits sum to 9. Now 9 = 4 + 5 = 3 + 6 = 2 + 7. Using 4 and 5 for a and d we obtain answer (v). Note that we can interchange the tens and hundreds column to give answer (vi).
Proceeding in this way we can look at 1062 (which leads to answer (vii) and (viii)) and 1260 (for which there is no answer, nor is there one for 1620). Hence we produce four answers.
Now turning to 1035 and its permutations we gather the remaining four answers.
If you can see a non-algebraic approach to this problem that doesn’t take for ever (see Solution to Jim’s problem above), we’d like to see it. A successful non-algebraic approach would be nice to put into this Solution. So would a more efficient way of doing the whole problem.
AttachmentSize
## Nine Tiles
Guess and check the solution to a number problem;
Explain the effect of "carry-overs" in addition problems.
Use the numbers facts with digits to solve a problem;
Use a systematic approach to get the number of ways that four numbers can add to 9.
| 3,720
| 11,793
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875
| 4
|
CC-MAIN-2013-20
|
longest
|
en
| 0.959379
|
https://math.stackexchange.com/questions/3089388/impulse-response-of-integrator
| 1,579,511,843,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250598217.23/warc/CC-MAIN-20200120081337-20200120105337-00380.warc.gz
| 543,070,042
| 30,752
|
# Impulse response of integrator
I want to get the impulse response of an LTI system where $$y(t) = \int_{t-2T}^{t-T} x(\alpha) d\alpha$$
To solve this I did: $$h(t) = \int_{t-2T}^{t-T} \delta(\alpha) d\alpha$$
Then you see that for the integrator to be around the impulse $$t > T$$ $$t < 2T$$
To me the impulse respsonse should be $$h(t)=\pi_{3T}(t-3T/2)$$ where pi is the rectangular function. Instead in the solution the answer is $$h(t)=\pi_{T}(t-3T/2)$$ According to me the subscript of the rectangular function is the width of it. How can this be T?
You showed that your impulse response is $$1$$ for
$$T < t < 2T$$
meaning that the length of its support is $$2T - T = T$$, and not $$T + 2T = 3T$$.
• So the rectangular function of length $T$ is centered at the origin. So you need to shift this to the middle of your support (i.e., $1.5T = (3/2)T$) – Metric Jan 27 '19 at 15:20
• One thing: By support I mean the interval $(T,2T)$, and the middle of this interval is $1.5T$, so we needed to shift the rectangular function that's centered at $0$ to the middle of the interval $(T,2T)$ (which is $1.5T$). The $T$ I referenced above is the length of the interval $(T,2T)$. When you see the term "support" in engineering, you can think of it as just the set of all values where the function is not zero [I should've clarified this] – Metric Jan 27 '19 at 19:18
| 435
| 1,368
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 10, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.75
| 4
|
CC-MAIN-2020-05
|
latest
|
en
| 0.870104
|
https://www.urionlinejudge.com.br/judge/en/profile/284593
| 1,611,596,698,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610703587074.70/warc/CC-MAIN-20210125154534-20210125184534-00306.warc.gz
| 1,022,620,912
| 6,007
|
# PROFILE
Check out all the problems this user has already solved.
Problem Problem Name Ranking Submission Language Runtime Submission Date
1914 Whose Turn Is It? 00295º 12421957 Java 0.064 12/2/18, 1:48:58 PM
1179 Array Fill IV 00276º 12413953 Java 0.036 12/1/18, 12:16:53 PM
1175 Array change I 00168º 12413889 Java 0.032 12/1/18, 12:06:45 PM
1182 Column in Array 00097º 12413843 Java 0.040 12/1/18, 12:01:30 PM
1173 Array fill I 00043º 12413806 Java 0.024 12/1/18, 11:51:28 AM
1172 Array Replacement I 00496º 12413771 Java 0.036 12/1/18, 11:46:49 AM
1181 Line in Array 00116º 12413747 Java 0.040 12/1/18, 11:42:29 AM
1183 Above the Main Diagonal 00150º 12364166 Java 0.044 11/26/18, 2:05:11 PM
1185 Above the Secundary Diagonal 00054º 12364153 Java 0.036 11/26/18, 2:04:17 PM
1180 Lowest Number and Position 00269º 12363110 Java 0.048 11/26/18, 12:18:56 PM
1157 Divisors I 00067º 12362914 Java 0.028 11/26/18, 11:56:11 AM
1153 Simple Factorial 00124º 12362876 Java 0.028 11/26/18, 11:42:31 AM
1132 Multiples of 13 00239º 12362772 Java 0.032 11/26/18, 11:41:03 AM
1188 Inferior Area 00075º 12362732 Java 0.040 11/26/18, 11:35:11 AM
1065 Even Between five Numbers 01380º 12421799 Java 0.044 11/26/18, 11:18:34 AM
1048 Salary Increase 00100º 12362550 Java 0.040 11/26/18, 11:13:21 AM
1044 Multiples 01221º 12362474 Java 0.048 11/26/18, 10:57:19 AM
1043 Triangle 00283º 12362377 Java 0.044 11/26/18, 10:52:20 AM
1847 Welcome to the Winter! 00043º 12362320 Java 0.040 11/26/18, 10:42:11 AM
1078 Multiplication Table 00168º 12316901 Java 0.028 11/21/18, 12:25:25 AM
1049 Animal 00834º 12316854 Java 0.048 11/21/18, 12:21:12 AM
1174 Array Selection I 00058º 12316818 Java 0.036 11/21/18, 12:17:50 AM
1189 Left Area 00023º 12316794 Java 0.032 11/21/18, 12:15:22 AM
1929 Triangle 00088º 12316763 Java 0.060 11/21/18, 12:10:45 AM
1042 Simple Sort 00059º 12316690 Java 0.032 11/21/18, 12:06:09 AM
1041 Coordinates of a Point 00946º 12301791 Java 0.052 11/19/18, 11:45:39 AM
1019 Time Conversion 00292º 12301539 Java 0.036 11/19/18, 11:17:48 AM
1828 Bazinga! 00091º 12301523 Java 0.044 11/19/18, 11:16:15 AM
1011 Sphere 01456º 12301507 Java 0.048 11/19/18, 11:14:14 AM
1001 Extremely Basic 08934º 12301475 Java 0.052 11/19/18, 10:53:49 AM
1 of 1
| 1,022
| 2,238
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.875
| 3
|
CC-MAIN-2021-04
|
latest
|
en
| 0.360479
|
http://math.stackexchange.com/questions/244617/why-is-every-solution-to-a-homogenous-second-order-linear-differential-equation
| 1,469,751,754,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-30/segments/1469257829320.91/warc/CC-MAIN-20160723071029-00217-ip-10-185-27-174.ec2.internal.warc.gz
| 167,449,836
| 18,367
|
# Why is every solution to a homogenous second-order linear differential equation in the form $C_0e^{\alpha x} + C_1e^{\beta x}$
In textbooks, it's often casually mentioned, without explanation, that any two solutions added together is the general solution, the form of every other solution.
I don't understand why this is or where that idea comes from. Can anyone explain why this is so?
EDIT: This question was badly put. I'm going to post another question that makes more sense. Thanks for the feedback.
-
Sometimes the general solution has the form $(C_0+C_1x)e^{\alpha x}$. – Hagen von Eitzen Nov 25 '12 at 23:37
It's not true that every solution has the form given in the title of the question.
It is true that any linear combination of solutions is a solution or, to put the same thing another way, that the set of all solutions forms a vector space. Do you need help proving that the set of all solutions forms a vector space?
The only question, then, is why the vector space has dimension 2. Well, one proves a theorem stating that there is a unique solution given the initial conditions $y(0)$ and $y'(0)$, and that settles it.
-
The premise in your title is incorrect, even for constant-coefficient equations (and certainly not for non-constant-coefficient ones).
For example, $y'' + 2 y' + y = 0$ has the solutions $y = C_0 e^{-x} + C_1 x e^{-x}$.
What is true is that by the Existence and Uniqueness Theorem and linearity, the space of solutions of a homogeneous second-order linear differential equation is a two-dimensional vector space. So if you have two linearly independent solutions, all solutions are linear combinations of these two.
-
While excellent answers have already been given, I think you might like some more concrete steps. Say I have two solutions, $y_1$ and $y_2$. If the equation is represented by some linear differential operator $D$ (it could be $D=\frac{d^2}{dx^2}+x$, $D=x^2\frac{d^2}{dx^2}-\frac{d}{dx}$ or any number of things), then
$$D(y_1)=0$$ $$D(y_2)=0$$
these are zero precisely because $y_1$ and $y_2$ are both solutions to the homogeneous ODE by assumption. So clearly
$$D(y_1)+D(y_2)=0$$
and if the operator is linear (which it is here)
$$D(y_1+y_2)=0$$
So if you have two solutions, by linearity their sum is a solution as well. At its core this is just a fancy application of $0+0=0$. A very similar statement can be made about scalar multiples. Why two independent solutions are needed to find the rest (i.e, they form a basis) has been answered excellently by others.
-
| 657
| 2,543
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2016-30
|
latest
|
en
| 0.938467
|
https://www.physicsforums.com/threads/the-dot-or-cross-product-of-two-operators-acting-on-a-state.754646/
| 1,508,670,850,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825174.90/warc/CC-MAIN-20171022094207-20171022114207-00243.warc.gz
| 978,181,900
| 14,927
|
# The dot or cross product of two operators acting on a state
1. May 21, 2014
### Robert_G
If a system is made up by two subsystems, for example, the atom and the photon. and let's assume the state of the atoms is described by $|\phi\rangle$, while the state of the photons can be described by $|n\rangle$, The Kronecker product of the $|\phi\rangle$ and $|n\rangle$ can be used to describe the whole system. and that would be:
$|\Psi\rangle=|\phi, n\rangle=|\phi\rangle\otimes|n\rangle$
I always treat the $|\phi\rangle$ and $|n\rangle$ as vectors, so the operation of $\otimes$ means the elements of the first vector (here $|\phi\rangle$) times the "whole" following vector which is $|n\rangle$ here; that will gives us a vector which is $|\phi, n\rangle$. so if the numbers of the elements of $|\phi\rangle$ and $|n\rangle$ is $m$ and $n$ respectively, the vector $|\phi, n\rangle$ has $m\times n$ elements.
now for example, we have two operators, $\hat{\mathbf{A}}$ and $\hat{\mathbf{N}}$, and they satisfy the following equations:
$\hat{\mathbf{A}}|\phi\rangle=\mathbf{a}|\phi'\rangle$
$\hat{\mathbf{N}}|n\rangle=\mathbf{n}|n'\rangle$.
Of course, $\hat{\mathbf{A}}$ can only act on the atomic states, and $\hat{\mathbf{N}}$ can only act on the photons states.
Now, my question, what is $\hat{\mathbf{A}}\cdot \hat{\mathbf{N}}|\phi, n\rangle$, and what is $\hat{\mathbf{A}}\times \hat{\mathbf{N}}|\phi, n\rangle$? The idea just not clear to me, if the operation $\otimes$ is involved.
if $\hat{U}=\hat{\mathbf{A}}\cdot \hat{\mathbf{N}}$, for example, how to write $\langle \phi, n|\;|U|^2\; |\phi', n'\rangle$ on the base of $|\phi\rangle$ and $|n\rangle$?
2. May 21, 2014
### strangerep
You need the tensor product of operators. (See the section "Tensor product of linear maps".)
3. May 21, 2014
### Robert_G
are you talking about a section of a book?
4. May 22, 2014
### micromass
Staff Emeritus
| 612
| 1,918
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125
| 4
|
CC-MAIN-2017-43
|
longest
|
en
| 0.772261
|
https://blogs.mathworks.com/cleve/category/programming/?from=jp
| 1,653,496,312,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00110.warc.gz
| 178,357,453
| 23,513
|
# Cleve’s Corner: Cleve Moler on Mathematics and ComputingScientific computing, math & more
Posts 1 - 10 of 16
# Digital Simulation of Rubik’s Cube with Qube2
This blog post describes Qube, my Rubik's Cube simulator. Source code for Qube is now available in one single file from this link: Qube_osf.m. I will also submit the code to the MATLAB Central File Exchange. As usual, I welcome any feedback.... 続きを読む >>
# Rubik’s Cube2
My recent obsession with cubes in this blog has led me back to Rubik's cube, perhaps the greatest mathematical puzzle of all time.... 続きを読む >>
# The Menger Sponge, Complement and Associate2
A few months ago, I had never heard of Karl Menger or the cube-like fractal that he invented. Now I have this long blog post about the sponge, and an extension that I am calling its "associate".... 続きを読む >>
# Anatomy of a Cube3
A cube is the familiar three-dimensional solid with eight vertices, six faces and twelve edges. I have been working with cubes recently in posts about both the Menger sponge fractal and the 4-by-4 matrix from computer graphics.... 続きを読む >>
# Quaternions3
Quaternions are generalizations of the complex numbers that have found applications in computer graphics and many other fields.... 続きを読む >>
# Solving Commodious Linear Systems2
This is about linear systems with fewer equations than variables; A*x = b where the m -by- n matrix A has fewer rows that columns, so m < n . I have always called such systems wide or fat, but this is not respectful. So I consulted the Merriam-Webster Thesaurus and found commodious.... 続きを読む >>
Posts 1 - 10 of 16
| 406
| 1,607
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.917201
|
https://www.mrexcel.com/board/threads/returing-a-vaule-based-on-the-first-letter-of-another-cell.805274/
| 1,726,708,799,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651944.55/warc/CC-MAIN-20240918233405-20240919023405-00420.warc.gz
| 819,722,968
| 19,797
|
# Returing a vaule based on the first letter of another cell
#### Petteri
##### New Member
Hello Everyone!
I'm working on a project where I need to return a set value (1,2,3 etc.) based on the first letter of another cell. However there is an additional wrinkle to this. The letters are organzined into certain groups. For example the letters A,B,C are always in one goup. I need to be able to define this group and then be able to set this group the set value.
Groups: (A,B,C) = 1, (D,E,F,G,H) = 2, (I, J,K,L) =3, (M, N, O) = 1 etc....
Example:
City Code To carousel AUS 1 BDL 1 BOG 1 BOS 1 CUN 1 DCA 1 EWR 1 HPN 1
<TBODY>
</TBODY>
I hope this makes sense, so I need the To Carousel value to fill in the correct value depeding on the first letter of the City Code column.
Thanks for any help!!
### Excel Facts
Test for Multiple Conditions in IF?
Use AND(test, test, test, test) or OR(test, test, test, ...) as the logical_test argument of IF.
Create a 2 column table of correspondences on a sheet named Admin like below...
A2: A, B, C
A3: D, E, F, G, H
...
B2: 1
B3: 2
...
Select this range and name the selection as CTable (from correspondence table).
Now invoke in B2 (under the header To carousel)...
Rich (BB code):
``````=LOOKUP(9.99999999999999E+307,S
EARCH(LEFT(A2),","&SUBSTITUTE(INDEX(CTable,0,1)," ","")),
INDEX(CTable,0,2))
``````
EDIT: my mistake there were some unwanted spaces in the formula. It's now working!!
Thank you so much for the answer. I'm very new at Excel. I've done the above but I'm getting a #NAME error.
Destination Number of flights Carousel Assigned AUS 1 #NAME? BDL 1 1 BOG 1 1 BOS 3 1 CUN 1 1
<TBODY>
</TBODY>
CTable:
A, B, C 1 D,E, F, G, H 1 I, J, K, L 2 M, N, O 1 P, Q, R 3 S 3 T, U, V, W, X, Y, Z 3
<TBODY>
</TBODY>
Formula used: =LOOKUP(9.99999999999999E+307,S
EARCH(LEFT(A2),","&SUBSTITUTE(INDEX(CTable,0,1)," ","")),
INDEX(CTable,0,2))
Thanks again! I'm not sure what I've done wrong....
Last edited:
@Petteri
You are welcome.
Ok, I've got one last question reguarding this sheet. If you don't mind, it's certainly appreciated!
I now need to total some values based on the carousel assigned:
Destination Number of flights Carousel Assigned Number of seats AUS 1 1 150 BDL 1 1 150 BOG 1 1 150 BOS 3 1 450 CUN 1 1 150 1050
<TBODY>
</TBODY><COLGROUP><COL><COL><COL><COL></COLGROUP>
Carousel 1 Daily Total 2750 Carousel 2 Daily Total 2630 Carousel 3 Daily Total 2150
<TBODY>
</TBODY><COLGROUP><COL><COL></COLGROUP>
I'm currently using this, but it's just a manual process to define the cells and won't auto update: =SUM(D7,D11, D22)
Have a look at the SUMIF function.
Replies
4
Views
3K
Replies
8
Views
991
Replies
0
Views
1K
Replies
9
Views
1K
Replies
9
Views
659
1,221,418
Messages
6,159,791
Members
451,589
Latest member
Harold14
### We've detected that you are using an adblocker.
We have a great community of people providing Excel help here, but the hosting costs are enormous. You can help keep this site running by allowing ads on MrExcel.com.
### Which adblocker are you using?
1)Click on the icon in the browser’s toolbar.
2)Click on the icon in the browser’s toolbar.
2)Click on the "Pause on this site" option.
Go back
1)Click on the icon in the browser’s toolbar.
2)Click on the toggle to disable it for "mrexcel.com".
Go back
### Disable uBlock Origin
Follow these easy steps to disable uBlock Origin
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
### Disable uBlock
Follow these easy steps to disable uBlock
1)Click on the icon in the browser’s toolbar.
2)Click on the "Power" button.
3)Click on the "Refresh" button.
Go back
| 1,152
| 3,700
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2024-38
|
latest
|
en
| 0.863454
|
https://www.physicsforums.com/threads/fortran-project.348028/
| 1,513,487,335,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948592972.60/warc/CC-MAIN-20171217035328-20171217061328-00114.warc.gz
| 792,160,482
| 15,752
|
# Fortran Project
1. Oct 22, 2009
### Physicsdude13
1. The problem statement, all variables and given/known data
Hi, I have to do a project in Fortran based on solving a system. My professor mentioned one idea to me, I am trying to see if this idea is even feasible and some potential ways to progress through it before I submit a project proposal to do it.
The problem is based on the square packing problem.
So for example a user inputs the coordinates of a large rectangle. The program then generates 100 squares all of varying sizes, and maximizes the amount of surface area the rectangle can be covered with those random squares generated. The squares cannot overlap or go outside the boundaries. Also, for extra credit possibly to show this as a GUI.
2. Relevant equations
This is an introductory course, so I may not have learned all the tools to fortran I need yet. But can anyone give me an idea as to the degree of programming difficulty. I know most of the basics, like subroutines, functions, loops, conditionals, variables, input, output.
3. The attempt at a solution
I am pretty stumped as to an algorithm to solve this as well. So far I can only think of doing the following pseudo-code:
1.User inputs 3-5 coordinates (putting a maximum at a 5 sided shape)
2.Program randomizes a variable amount of squares of random sizes within a range (the amount of squares generated can be determined by user, that way to increase the amount of surface area covered you just generate more squares sort of like an approximation method for the surface area)
3.Potentially represent the squares generated as coordinates as well in array format?
From 3 on, I am stumped: Any feedback would be much appreciated. Thank you.
| 379
| 1,732
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.703125
| 3
|
CC-MAIN-2017-51
|
longest
|
en
| 0.946075
|
http://guiartys.com/yuroc.php?e34854=electricity-and-magnetism-are-different-because-quizlet
| 1,603,553,360,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107883636.39/warc/CC-MAIN-20201024135444-20201024165444-00100.warc.gz
| 51,409,259
| 6,400
|
# electricity and magnetism are different because quizlet
It's interesting that static electricity also acts at a distance, and isn't hard to observe, but it doesn't seem to get swept into the same mental category as gravity and magnetism.
Unit Overview Electricity can be exciting and fun. This 6.7 Describe how electricity can be helpful or harmful to people (safety).
The magnetic field is the force from magnetic material. The tape pieces repel each other because they have picked up a net charge, hence there is an electric field between ... place the test mouse at different points in the room with the cat and the comb as shown below. The electric field is a natural force for charged particles. One of the best ways to save electricity is to put off unnecessary lights. Although conceived of as distinct phenomena until the 19th century, electricity and magnetism are now known to be components of the unified theory of electromagnetism. The electricity is the flow of charge per second. properties of different materials.
We are familiar with in our everyday lives with the phenomenon of static cling – when two objects, such as a piece of Saran wrap and a wool sweater, are rubbed together, they cling. 6.5 Investigate properties of magnet s. 6.6 Explore the interaction of electricity and magnetism to create an electromagnet. Electricity and magnetism are ultimately inextricably linked. Electricity and Magnetism Electricity and magnetism is an interesting aspect of electricity.
Maybe that's somehow because we usually have to do something active to generate the … There are many ways to save the cost of electricity. Conserving electricity helps save a lot of money and also helps to reduce global warming. Similar: You have a force from one polarity to another. Make use of more natural light. A connection between electricity and magnetism had long been suspected, and in 1820 the Danish physicist Hans Christian Orsted showed that an electric current flowing in a wire produces its own magnetic field. Electricity and Magnetism | SEE Physics Notes Electricity: Definition: The flow of electrons through conductor is called Electricity. Electricity is the term given to a group of physical phenomena involving electric charges, their motions, and their effects.
Q = 2t or, i = Q/t = charge/ time Father of Electricity: Michael Faraday Types of electricity: a) Ac (Alternating current) b) DC (Direct current), etc. It can make your hair stand on end (static electricity).
| 501
| 2,493
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2020-45
|
longest
|
en
| 0.953702
|
https://codefinity.com/courses/v2/671389bc-34ed-4de7-83cd-2d1bfcf00a76/7c4e6464-e82b-4f01-bcb9-78d53d060da5/3e103e96-80ec-4eb6-88ce-8ad518238b58
| 1,716,793,208,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059037.23/warc/CC-MAIN-20240527052359-20240527082359-00135.warc.gz
| 136,712,781
| 45,500
|
Notice: This page requires JavaScript to function properly.
Please enable JavaScript in your browser settings or update your browser.
Function array()
Course Content
# NumPy in a Nutshell
NumPy in a Nutshell
## Function array()
In fact, there are various functions in NumPy for creating arrays. Now, we'll explore one of the most commonly used ones, namely `np.array()`. Below, you'll find an example of how to use this function:
Let's now determine the type of object that this function creates. We can do this using the well-known function `type()`.
Note
The `type()` function takes an object of any type and returns its type. The argument can indeed be of any type: number, string, list, dictionary, tuple, function, class, module, etc.
We can see the type of the created array is `ndarray`. But what does that mean? ndarray - This object is a multidimensional homogeneous array with a predetermined number of elements.
Now it's time to practice!
1. You have to create two NumPy arrays. The first one should look like this: `[65, 2, 89, 5, 0, 1]` and the second one should look like this: `[1, 2, 3]`.
| 267
| 1,115
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.765625
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.789606
|
https://www.airmilescalculator.com/distance/geg-to-ase/
| 1,623,752,450,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00431.warc.gz
| 570,808,042
| 46,226
|
# Distance between Spokane, WA (GEG) and Aspen, CO (ASE)
Flight distance from Spokane to Aspen (Spokane International Airport – Aspen–Pitkin County Airport) is 788 miles / 1269 kilometers / 685 nautical miles. Estimated flight time is 1 hour 59 minutes.
Driving distance from Spokane (GEG) to Aspen (ASE) is 1130 miles / 1819 kilometers and travel time by car is about 19 hours 12 minutes.
## Map of flight path and driving directions from Spokane to Aspen.
Shortest flight path between Spokane International Airport (GEG) and Aspen–Pitkin County Airport (ASE).
## How far is Aspen from Spokane?
There are several ways to calculate distances between Spokane and Aspen. Here are two common methods:
Vincenty's formula (applied above)
• 788.488 miles
• 1268.949 kilometers
• 685.178 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth’s surface, using an ellipsoidal model of the earth.
Haversine formula
• 787.874 miles
• 1267.961 kilometers
• 684.644 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## Airport information
A Spokane International Airport
City: Spokane, WA
Country: United States
IATA Code: GEG
ICAO Code: KGEG
Coordinates: 47°37′11″N, 117°32′2″W
B Aspen–Pitkin County Airport
City: Aspen, CO
Country: United States
IATA Code: ASE
ICAO Code: KASE
Coordinates: 39°13′23″N, 106°52′8″W
## Time difference and current local times
The time difference between Spokane and Aspen is 1 hour. Aspen is 1 hour ahead of Spokane.
PDT
MDT
## Carbon dioxide emissions
Estimated CO2 emissions per passenger is 134 kg (295 pounds).
## Frequent Flyer Miles Calculator
Spokane (GEG) → Aspen (ASE).
Distance:
788
Elite level bonus:
0
Booking class bonus:
0
### In total
Total frequent flyer miles:
788
Round trip?
| 493
| 1,928
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2021-25
|
latest
|
en
| 0.828668
|
https://www.mathtestpreparation.com/Principles/QuadraEllipse.aspx
| 1,656,993,742,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104512702.80/warc/CC-MAIN-20220705022909-20220705052909-00146.warc.gz
| 937,422,492
| 2,042
|
back to math help
Ellipse equation:
x2/a2 + y2/b2= 1
( a > b > 0 )
where c2 = a2 - b2
Center(0, 0)
foci (-c, 0) and (c, 0)
Ellipse equation:
(x - h)2/a2 + (y - k)2/b2 = 1
( a > b > 0 )
where c2 = a2 - b2
Center(h, k)
foci (h-c, h+c) and (h+c, h-c)
Ellipse equation:
y2/a2 + x2/b2= 1
( a > b > 0 )
where c2 = a2 - b2
Center(0, 0)
foci (0, -c) and (0, c)
x2/225 + y2/100 = 1
Center(0, 0)
F(-11.18, 0) and F(11.18, 0)
(x - 5)2/196 + y2/100 = 1
Center(5, 0)
F(-4.8, 0) and F(14.8, 0)
(x + 5)2/100 + (y + 5)2/64 = 1
Center(-5, -5)
F(-11, -5) and F(1, -5)
x2/100 + (y - 5)2/196 = 1
Center(0, 5)
F(0, 14.8) and F(0, -4.8)
(x + 5)2/100 + y2/64 = 1
Center(-5, 0)
F(-11, 0) and F(1, 0)
| 389
| 676
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2022-27
|
latest
|
en
| 0.47654
|
https://repl.it/talk/share/ChezCoder-Message-the-Hoot-Boot-bot-i/38527/174283
| 1,590,843,140,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347409171.27/warc/CC-MAIN-20200530102741-20200530132741-00097.warc.gz
| 524,858,728
| 58,950
|
The L Programming Language
# The L Programming Language
This is a programming language that I made as my final project for my CS class this past year (for those of you who are curious, I got a 98% on it). It ended up turning out much better than I had expected, so I thought it'd be cool to share it here on Repl Talk.
## What is L?
L is an array-oriented programming language made for performing complex operations on large sets of data. It is inspired by languages of the same category such as R, Julia, and J (which was my main inspiration).
While it appears to be an abstract mess of symbols at first, the conciseness of using symbolic operators as opposed to named functions allows you to do more by doing less.
J was my main inspiration for L. I had found J a few months ago after doing some stuff for the APL language mode here on repl.it, mainly because J is highly influenced by APL (minus its Unicode character set).
The core concept in both languages is that you can create new functions by composing existing functions, all of which can either be a unary operator (monadic; takes one value) or a binary operator (dyadic; takes 2 values). While L does not allow you to compose functions (that would have taken way more work and I didn’t have time for that anyways), it does have the monadic and dyad operator features.
Another core concept is that when you perform operations on an array of values, the operation is performed on each value of the array. For example, doing `1 2 3 + 1` will give you `2 3 4` as a result. If you perform a binary operation on 2 arrays of the same length, it will perform the operation on each element of the arrays in parallel. For example, doing `1 2 3 + 4 5 6` will give you `5 7 9`.
## If I had more time…
If I had more time to work on this, I would have added things like:
• More mathematical operators (log, trig, etc).
• More statistical operators (min, max, average, etc).
• More operators in general.
• “Boxed” sets (that don’t perform operations on every value it has).
• “Flat” sets (opposite of above but one level deeper (if that makes sense)).
• User-defined functions/operators.
• “Lazy” expressions that can be assigned to variables and aren’t evaluated until they’re referenced/used.
• Better documentation.
## Documentation
Because the original documentation for my project is super long (too long for a Repl Talk post), I've included it in the repl files as a PDF (file name is `Docs.pdf`).
## Examples
### Factorial
``````num: 5
{* (1..num)``````
### Filter numbers between (min, max]
``````range: 1..10
min: 2
max: 8
range ?> min ?<= max``````
### Prime number checker
``````num: 17
#(num / (2..num-1) % 1 ?= 0) = 0``````
## Future thoughts
Might try adding more features to this in the future like user-defined functions or something. Also might try rewriting this in Rust since it'd have less boilerplate than doing it like this in C++
You are viewing a single comment. View All
theangryepicbanana (1530)
@ChezCoder Message the "Hoot Boot" bot if you'd like to contact the mods privately then :)
| 761
| 3,083
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.5625
| 3
|
CC-MAIN-2020-24
|
latest
|
en
| 0.963332
|
https://artofproblemsolving.com/wiki/index.php?title=Greatest_common_divisor&diff=prev&oldid=60731
| 1,607,025,438,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141732696.67/warc/CC-MAIN-20201203190021-20201203220021-00109.warc.gz
| 192,989,602
| 10,995
|
# Difference between revisions of "Greatest common divisor"
The greatest common divisor (GCD, or GCF (greatest common factor)) of two or more integers is the largest integer that is a divisor of all the given numbers.
The GCD is sometimes called the greatest common factor (GCF).
A very useful property of the GCD is that it can be represented as a sum of the given numbers with integer coefficients. From here it immediately follows that the greatest common divisor of several numbers is divisible by any other common divisor of these numbers.
## Finding the GCD
### Using prime factorization
Once the prime factorizations of the given numbers have been found, the greatest common divisor is the product of all common factors of the numbers.
Example:
$270=2\cdot3^3\cdot5$ and $144=2^4\cdot3^2$. The common factors are 2 and $3^2$, so $GCD(720,144)=2\cdot3^2=18$.
### Euclidean algorithm
The Euclidean algorithm is much faster and can be used to give the GCD of any two numbers without knowing their prime factorizations. To find the greatest common divisor of more than two numbers, one can use the recursive formula $GCD(a_1,\dots,a_n)=GCD(GCD(a_1,\dots,a_{n-1}),a_n)$.
### Using least common multiple
The GCD of two numbers can also be found using the equation $GCD(x, y) \cdot LCM(x, y) = x \cdot y$, where $LCM(x,y)$ is the least common multiple of $x$ and $y$.
| 350
| 1,379
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125
| 4
|
CC-MAIN-2020-50
|
latest
|
en
| 0.918555
|
https://www.queryhome.com/puzzle/7427/staircase-steps-you-walk-taking-steps-time-how-many-ways-stair
| 1,726,836,665,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700652278.82/warc/CC-MAIN-20240920122604-20240920152604-00296.warc.gz
| 873,121,112
| 26,742
|
# A staircase has 10 steps , you can walk up taking one or two steps at a time . how many ways can go up to top of stair ?
851 views
A staircase has 10 steps , you can walk up taking one or two steps at a time . how many ways can go up to top of stair ?
posted May 14, 2015
10C0 + 9C1 + 8C2 + 7C3 + 6C4 + 5C5
= 1+9+28+35+15+1
= 89
Sir , i didn't got your solution ,
my solution is of fibonacci series , like for 1step ==1 , 2nd step 1+1 or 2 ==2ways , 3rd step 1+1+1 or 1+2 or 2+1 ==3ways , so ,1,2,3,5,8,13,21,34,55,89 , for 10th step 89 ways .
I just tried for two three numbers and saw this logic is arriving (not sure why), but fibonacci series is more logical :) thanks for asking great puzzle.
I am glad to be of help :)
Similar Puzzles
You need to climb ten stairs. At every stair, you can either take one step up or you can jump two steps up.
In how many different ways you can climb 10 stairs?
+1 vote
Garima takes the underground train to work and uses an escalator at the railway station. If Garima runs up 8 steps of the escalator, then it takes her 30.0 seconds to reach the top of the escalator. If she runs up 15 steps of the escalator, then it takes her only 12.5 seconds to reach the top.
How many seconds would it take Garima to reach the top if she did not run up any steps of the escalator at all?
| 411
| 1,329
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2024-38
|
latest
|
en
| 0.929661
|
https://discourse.julialang.org/t/generic-n-dimensional-loop-without-allocations/674
| 1,550,672,223,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-09/segments/1550247495001.53/warc/CC-MAIN-20190220125916-20190220151916-00284.warc.gz
| 548,278,200
| 8,574
|
# Generic n-dimensional loop without allocations
#1
I’m working on some code that should iterate over milti-dimensional space, where the dimension `dim` is provided as a parameter to the function. The simplest approach I found is to use CartesianRange, like in the following simple example:
``````function test1(dim::Int)
imin=fill(0, dim)
imax=fill(0, dim)
accumulator = 0
for k = 1:10000
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in CartesianRange(CartesianIndex(imin...), CartesianIndex(imax...))
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
return accumulator
end
``````
While this works great, there is a finite memory allocation for each splatting imin… and imax…
``````@time test1(2);
0.191940 seconds (179.32 k allocations: 5.626 MB)
``````
For the real application, this alone can exhaust my available memory, as many such multidimensional loops are required.
A very similar function with “hard-coded” dim=2 completely avoids this problem, and is also much faster
``````function test2()
imin=fill(0, 2)
imax=fill(0, 2)
accumulator = 0
for k = 1:10000
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in CartesianRange(CartesianIndex(imin[1],imin[2]), CartesianIndex(imax[1],imax[2]))
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
return accumulator
end
``````
`````` 0.001797 seconds (7 allocations: 368 bytes)
``````
Now the question: is it possible to avoid this multidimensional loop allocation without sacrificing hard-coding the dimensions and the splatting of the CartesianRange?
Cheers!
#2
I figured out a possible solution using a generated function
``````@generated function build_iterator{N}(dim::Type{Val{N}}, imin::Vector{Int}, imax::Vector{Int})
if dim==Type{Val{1}}
return :(CartesianRange(CartesianIndex(imin[1]), CartesianIndex(imax[1])))
elseif dim==Type{Val{2}}
return :(CartesianRange(CartesianIndex(imin[1],imin[2]), CartesianIndex(imax[1],imax[2])))
elseif dim==Type{Val{3}}
return :(CartesianRange(CartesianIndex(imin[1],imin[2], imin[3]), CartesianIndex(imax[1],imax[2], imax[3])))
else
return :(CartesianRange(CartesianIndex(imin...), CartesianIndex(imax...)))
end
end
test3(dim::Int) = test(dim, Val{dim})
function test3{N}(dim::Int, ::Type{Val{N}})
imin=fill(0, N)
imax=fill(0, N)
accumulator = 0
for k = 1:10000
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in build_iterator(Val{N}, imin, imax)
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
accumulator
end
``````
Now this is just as fast as the hard-coded version
``````@time test3(2)
0.001346 seconds (7 allocations: 368 bytes)
``````
Please do comment if you know of a simpler or more elegant solution!
#3
you could probably also use StaticArrays, which are essentially tuples, to avoid allocation of arrays, then you avoid generated functions.
#4
For the elegance part of the question, you may avoid the `if` cases by utilizing something like this:
``````ex = :(imin[1])
for i in 2:(dim.parameters[1].parameters[1])
ex = :(\$ex, imin[\$i])
end
``````
#5
Interesting, thanks @cortner! Your suggestion is to use a StaticArray for the ranges imin and imax, correct? The problem I see is that StaticArrays cannot be mutated in-place, can they? So how could one change imin and imax in the above example (the rand! lines) without allocating a new StaticArray at each cycle?
#6
Haha, that’s very cool @akis, cheers.
``````@generated function build_iterator{N}(dim::Type{Val{N}}, imin::Vector{Int}, imax::Vector{Int})
exmin = :(imin[1])
exmax = :(imax[1])
for i in 2:(dim.parameters[1].parameters[1])
exmin = :(\$exmin, imin[\$i])
exmax = :(\$exmax, imax[\$i])
end
return :(CartesianRange(CartesianIndex(\$exmin), CartesianIndex(\$exmax)))
end
``````
#7
StaticArrays provides the MVector type, which is mutable but still fixed-size, so you can use its length as part of your function signature:
``````function foo{N, T}(imin::MVector{N, T})
...
``````
Also, it’s worth noting that constructing a new `SVector` at every loop iteration might actually be even faster, because constructing a new `SVector` does not actually allocate any heap memory (it just lives on the stack).
#8
Sorry that is what I meant - SVecror allocation is cheap.
#9
I tried the StaticArray approach, but I cannot get it to perform as well as the generated function approach. I tried both SVector and MVector. This is the mutating version
``````using StaticArrays
function test4(dim::Int)
imin=zeros(MVector{dim, Int})
imax=zeros(MVector{dim, Int})
accumulator = 0
for k = 1:10000
# In actual code, imin and imax are the result of a non-trivial calculation
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in CartesianRange(CartesianIndex(imin.data), CartesianIndex(imax.data))
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
accumulator
end
``````
This is the static version. I cannot get around the need to have some mutating variable at some point in the real code, because the ranges imin and imax are the result of a non-trivial calculation.
``````using StaticArrays
function test5(dim::Int)
imin=fill(0, dim)
imax=fill(0, dim)
accumulator = 0
for k = 1:10000
# In actual code, imin and imax are the result of a non-trivial calculation
rand!(imin, 1:5)
rand!(imax, 1:5)
smin = SVector{dim}(imin)
smax = SVector{dim}(imax)
for c in CartesianRange(CartesianIndex(smin.data), CartesianIndex(smax.data))
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
accumulator
end
``````
Both these tests perform as badly as test1 above. I’m most probably doing something wrong here
``````@time test4(2)
0.208932 seconds (193.47 k allocations: 6.058 MB)
``````
``````@time test5(2)
0.370437 seconds (377.44 k allocations: 16.555 MB, 1.80% gc time)
``````
#10
The `build_iterator` function can be written more simply using the macros from `Base.Cartesian`, such as `@ntuple` or `@ncall`. For example
``````@generated function build_iterator{N}(dim::Type{Val{N}}, imin, imax)
quote
CartesianRange(
CartesianIndex(@ntuple \$N n -> imin[n]),
CartesianIndex(@ntuple \$N n -> imax[n]),
)
end
end
``````
with `test_6` defined as
``````function test_6{N}(dim::Type{Val{N}})
imin = zeros(Int, N)
imax = zeros(Int, N)
acc = 0
for k = 1:10_000
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in build_iterator(dim, imin, imax)
acc += c.I[1] + c.I[2]
end
end
return acc
end
``````
I get
``````julia> @time test_6(Val{4})
0.001851 seconds (7 allocations: 400 bytes)
228336
``````
For an approach that doesn’t used any `@generated` trickery I’d go with something like
``````function test_7{N}(dim::Type{Val{N}})
v_min = zeros(Int, N)
v_max = zeros(Int, N)
acc = 0
for k = 1:10_000
rand!(v_min, 1:5)
rand!(v_max, 1:5)
t_min = ntuple(n -> v_min[n], dim)
t_max = ntuple(n -> v_max[n], dim)
for c in CartesianRange(CartesianIndex(t_min), CartesianIndex(t_max))
acc += c.I[1] + c.I[2]
end
end
return acc
end
``````
with time
``````
julia> @time test_7(Val{4})
0.002572 seconds (20.01 k allocations: 312.891 KB)
229552
``````
For the `StaticArrays` examples the main problem that I can see is that `dim` is being passed to `test4` as a value rather than a type parameter so `imin` and `imax` won’t be type stable. You can check this with `@code_warntype test4(4)` and you’ll see a lot of read `Any`s all over the place. Changing the `test4` signature to `test4{dim}(::Type{Val{dim}})` should fix that I think.
Edit
Here’s a third option that doesn’t do any unnecessary extra allocations
``````randtuple(T, t) = _randtuple(T, t...)
_randtuple(T) = ()
_randtuple(T, h, t...) = (rand(T), _randtuple(T, t...)...)
function test_8(t_min, t_max)
acc = 0
for k = 1:10_000
t_min = randtuple(1:5, t_min)
t_max = randtuple(1:5, t_max)
for c in CartesianRange(CartesianIndex(t_min), CartesianIndex(t_max))
acc += c.I[1] + c.I[2]
end
end
return acc
end
test_8(dim) = test_8(dim, dim)
``````
It does need to be passed an initial tuple, but that could also probably be abstracted away as well:
``````julia> @time test_8((0,0,0,0,0,0,0,0))
0.016166 seconds (6 allocations: 256 bytes)
733783
``````
#11
Yes!! Thanks @mike, that’s it. I should have checked the `@code_warntype` output of course.
Properly type-stable versions of the SVector and MVector approaches are now as performant as the `@generated` approach. I put them all down here as a summary
``````using Base.Cartesian
@generated function build_iterator{N}(dim::Type{Val{N}}, imin, imax)
quote
CartesianRange(
CartesianIndex(@ntuple \$N n -> imin[n]),
CartesianIndex(@ntuple \$N n -> imax[n]),
)
end
end
test3(dim::Int) = test3(Val{dim})
function test3{N}(dim::Type{Val{N}})
imin=fill(0, N)
imax=fill(0, N)
accumulator = 0
for k = 1:10000
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in build_iterator(Val{N}, imin, imax)
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
accumulator
end
``````
``````using StaticArrays
test4(dim::Int) = test4(Val{dim})
function test4{N}(dim::Type{Val{N}})
imin=zeros(MVector{N, Int})
imax=zeros(MVector{N, Int})
accumulator = 0
for k = 1:10_000
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in CartesianRange(CartesianIndex(imin.data), CartesianIndex(imax.data))
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
accumulator
end
``````
``````using StaticArrays
test5(dim::Int) = test5(Val{dim})
function test5{N}(dim::Type{Val{N}})
imin=fill(0, N)
imax=fill(0, N)
accumulator = 0
for k = 1:10000
# In actual code, imin and imax are the result of a non-trivial calculation
rand!(imin, 1:5)
rand!(imax, 1:5)
for c in CartesianRange(
CartesianIndex(SVector{N}(imin).data),
CartesianIndex(SVector{N}(imax).data))
# do something with c, e.g
accumulator += c.I[1]+c.I[2]
end
end
accumulator
end
``````
Timing tests point to MVector (test4) as the best, although by a small margin. I also think it is the coolest one. I need to think a bit more about @mike’s test_8 idea.
``````@benchmark test3(2)
#[...]
memory estimate: 208.00 bytes
allocs estimate: 3
mean time: 1.327 ms (0.00% GC)
#[...]
``````
``````@benchmark test4(2)
#[...]
memory estimate: 80.00 bytes
allocs estimate: 3
mean time: 1.201 ms (0.00% GC)
#[...]
``````
``````@benchmark test5(2)
#[...]
memory estimate: 208.00 bytes
allocs estimate: 3
mean time: 1.538 ms (0.00% GC)
#[...]
``````
Edit: added test_7 and test_8 benchmarks for completeness
Edit 2: test_7 has a typo I think. dim should read N in the ntuple lines. Benchmark for the corrected version.
Edit 3: reverted edit2 (not a typo!)
``````@benchmark test_7(Val{2})
#[...]
memory estimate: 312.70 kb
allocs estimate: 20003
mean time: 1.586 ms (1.36% GC)
#[...]
``````
``````@benchmark test_8((0,0))
#[...]
memory estimate: 0.00 bytes
allocs estimate: 0
mean time: 1.801 ms (0.00% GC)
#[...]
``````
#12
Well I suppose if you’re looking for zero allocations then `test_8` does win there , pity about the slight slowdown though. For me the timings are slightly closer between the two:
``````mean time: 1.266 ms (0.00% GC) # test_8
mean time: 1.043 ms (0.00% GC) # test4
``````
perhaps due to Julia version differences/hardware.
Wasn’t a typo, though it does look like one at first. `ntuple` can take either an `Int` or a `Val{N}` where `N` is some integer. One is inferrable while the other one isn’t, hence the huge slow down in the edited benchmark. You can see the difference in behaviour with `@code_warntype`
``````julia> @code_warntype ntuple(x -> 1, Val{3}) # good
...
julia> @code_warntype ntuple(x -> 1, 3) # bad
...
``````
#13
Indeed
Regarding the ntuple thing, gotit! Thanks.
#14
Another non-allocating option, close to the `test_8` that @mike proposed, but that uses Val{N} would be
``````function test_9{N}(::Type{Val{N}})
acc = 0
for k = 1:10_000
t_min = ntuple(_->rand(1:5), Val{N})
t_max = ntuple(_->rand(1:5), Val{N})
for c in CartesianRange(CartesianIndex(t_min), CartesianIndex(t_max))
acc += c.I[1] + c.I[2]
end
end
return acc
end
``````
#15
Hey, wow, that’s nice!
``````@benchmark test_9(Val{2})
memory estimate: 0.00 bytes
allocs estimate: 0
mean time: 1.809 ms (0.00% GC)
``````
Lots to ruminate here for me… I still need to understand whether this test_9 idea can actually be applied into my real-use code, where the rand becomes a more complicated thing. It would also be nice to understand why, despite making zero allocations, this is still slower than the MVector way… Thanks a lot to everyone for the great ideas, by the way!
#16
Yeah, @pabloferz’s `ntuple` is good too, could also use `map(_ -> rand(1:5), t_min)` to get pretty much the same effect as `randtuple` in `test_8`. All those approaches probably compile down to quite similar code in the end I’d suspect.
From what I can tell from the output of `@profile` for both calls (`test_8` and `test4`) it seems that the `MVector` approach is making use of a different code path in `random.jl` that uses `setindex!` on the vector rather than “rebuilding” a completely new tuple each time with `ntuple` and `_ntuple`. Tuples are immutable, hence the difference in how the new set of random integers gets created. It seems that `StaticArrays` does use a tuple as the backing data behind `MVector`, but then does this to change elements in the tuple.
That’s just my guess from looking into the `StaticArrays` code though. Perhaps @andyferris will have some further insight into the differences.
| 4,210
| 13,329
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.1875
| 3
|
CC-MAIN-2019-09
|
latest
|
en
| 0.584534
|
https://heredragonsabound.blogspot.com/2016/12/how-to-decorate-mountain-part-2.html
| 1,527,016,296,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794864872.17/warc/CC-MAIN-20180522190147-20180522210147-00245.warc.gz
| 591,072,265
| 18,820
|
## Tuesday, December 13, 2016
### How to Decorate a Mountain, Part 2
In the previous post, I cleaned up the algorithm for drawing mountains a bit, and then I added some extra details / decorations to the drawings. In this posting, I'm going to continue on with adding details.
Looking back at my hand-drawn samples for inspiration, I notice that one style features scribbles on the mountain face:
These scribbles suggest extra texture or features and help to break up the empty area of the mountain face.
One immediate challenge in adding scribbles is just finding a spot on the mountain face. Just picking a random spot inside a polygon is non-trivial. Fortunately I'm not too worried about performance, so I'm free to use inefficient but easier-to-implement algorithms. My approach to finding a a random spot inside the face of the mountain is to construct a bounding box for the mountain face:
And then randomly select points within the bounding box until I get one that is also inside the mountain face.
This works but is not generally a recommended algorithm because it can take an indeterminate amount of time to find a point inside the polygon. But in my case roughly 50% of the points are going to be inside the polygon, so I'm willing to take that risk.
(ADDENDUM: After implementing this and using it for a while, I realized I was rarely starting scribbles up near the top of the mountain. I didn't like that much, so I switched how I pick the starting point for a scribble. Now, I draw a line from the peak of the mountain to a random spot on the right side of the mountain, like these examples:
Hey, maybe I should implement volcanoes... Uh, anyway, I then pick a random spot on the line as a starting point for the scribble. As it turns out, this fits in well with the next step.)
Now that I have a point, I want to identify the area where I'm going to draw the scribble. To do this, I project a line from the peak down through the point and create a box where the scribble will fall.
Projecting a line down from the peak ensures that the box for the scribble will have the right slope for it's placement on the mountains -- if the scribble is near to the right side of the mountain it will be sloped about like the right side of the mountain; if it is near the center of the mountain it will be close to straight up and down.
Once I've established the box, I need to check to see if it goes outside the mountain or if it interferes with any of the existing decorations. In the example above, it does both. When this happens, I throw the box and the point away, generate a new starting point and try again. (Again, this can take an indeterminate amount of time -- possibly forever if it isn't possible to position a scribble. So the algorithm has to have a guard on it that gives up entirely after a few hundred attempts.)
If there isn't a problem, then I can draw the scribble. To do the drawing, I'm using a different approach than with scribbling in the shaded side of the mountain. In this case, I don't have to fully fill the area, and I also want this to look different than the shading. So I'm just taking a line and perturbing it side-to-side.
Here are some more examples, with various size scribbles:
My intention is to use these face scribbles as an accent, but it turns out to look interesting if you put a lot of scribbles on each mountain:
And maybe even more interesting at map scale:
The scribbles give the mountains a "textured" look. I'm not sure this is my favorite style, but in some ways it looks pretty good.
Another option I'd like to add is to have more flat-topped mountains. Currently, every mountain starts off as an upside-down V. Both sides do get perturbed, so occasionally that results in mountains with round or flat tops:
But I'd like to be able to intentionally generate mountains with flat tops. Then (for example) I could have an area of the map with "old" mountains that are generally flattened, and another area with "new" mountains that are sharper.
To get flattened mountains, I need to insert a flat section between the two sides of the mountain. Hopefully the perturbation and smooth line drawing routines should turn that into some interesting shapes.
The look of the mountains is also dependent on the curve algorithm I use to draw them, so I can relax that to get mountains that look more rounded.
In the next posting I'll talk about how I combine multiple mountains.
#### 1 comment:
1. You're the Bob Ross of procedural scribbling!
| 1,017
| 4,525
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.21875
| 3
|
CC-MAIN-2018-22
|
latest
|
en
| 0.930693
|
https://www.geeksforgeeks.org/numpy-expm1-python/?ref=rp
| 1,653,013,596,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662530553.34/warc/CC-MAIN-20220519235259-20220520025259-00713.warc.gz
| 873,750,338
| 24,074
|
# numpy.expm1() in Python
• Last Updated : 29 Nov, 2018
numpy.expm1(array, out = None, where = True, casting = ‘same_kind’, order = ‘K’, dtype = None) :
This mathematical function helps user to calculate exponential of all the elements subtracting 1 from all the input array elements.
Parameters :
```array : [array_like]Input array or object whose elements, we need to test.
out : [ndarray, optional]Output array with same dimensions as Input array,
placed with result.
**kwargs : allows you to pass keyword variable length of argument to a function.
It is used when we want to handle named argument in a function.
where : [array_like, optional]True value means to calculate the universal
functions(ufunc) at that position, False value means to leave the
value in the output alone.
```
Return :
```An array with exponential(all elements of input array) - 1.
```
Code 1 : Working
`# Python program explaining``# expm1() function`` ` `import` `numpy as np`` ` `in_array ``=` `[``1``, ``3``, ``5``]``print` `(``"Input array : \n"``, in_array)`` ` `exp_values ``=` `np.exp(in_array)``print` `(``"\nExponential value of array element : "`` ``"\n"``, exp_values)`` ` `expm1_values ``=` `np.expm1(in_array)``print` `(``"\n(Exponential value of array element) - (1) "`` ``": \n"``, expm1_values)`
Output :
```Input array :
[1, 3, 5]
Exponential value of array element :
[ 2.71828183 20.08553692 148.4131591 ]
(Exponential value of array element) - (1) :
[ 1.71828183 19.08553692 147.4131591 ]
```
Code 2 : Graphical representation
`# Python program showing``# Graphical representation of ``# expm1() function`` ` `import` `numpy as np``import` `matplotlib.pyplot as plt`` ` `in_array ``=` `[``1``, ``1.2``, ``1.4``, ``1.6``, ``1.8``, ``2``]``out_array ``=` `np.expm1(in_array)`` ` `print``(``"out_array : "``, out_array)`` ` `y ``=` `[``1``, ``1.2``, ``1.4``, ``1.6``, ``1.8``, ``2``]``plt.plot(in_array, y, color ``=` `'blue'``, marker ``=` `"*"``)`` ` `# red for numpy.expm1()``plt.plot(out_array, y, color ``=` `'red'``, marker ``=` `"o"``)``plt.title(``"numpy.expm1()"``)``plt.xlabel(``"X"``)``plt.ylabel(``"Y"``)``plt.show() `
Output :
out_array : [ 1.71828183 2.32011692 3.05519997 3.95303242 5.04964746 6.3890561 ]
My Personal Notes arrow_drop_up
| 752
| 2,298
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.9375
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.514134
|
https://bsci-ch.org/how-many-ounces-is-24-grams/
| 1,653,352,160,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662562106.58/warc/CC-MAIN-20220523224456-20220524014456-00732.warc.gz
| 193,367,651
| 5,297
|
What is 24g in oz? transform 24 Grams (24g) come Ounces (oz) and show formula, brief background on the units and quick maths because that the conversion.
You are watching: How many ounces is 24 grams
## Enter Grams to transform to Ounces
### What is the Gram?
The gram is a unit of mass (acceptable for usage as weight on Earth) and is a multiple of one SI base unit with the price g.
Although there is no a prefix, it actually represents 1/1000 kg. The is the very first subdivision of the SI basic unit; kilogram and 100 g amounts to 3.527396195 ounces.
It is the most common unit because that measuring ingredient (except liquid) in cooking and purchasing food items in the people today.
The bulk of nutritional values and information is express in regards to "per 100g".
1 g is roughly equal come the weight of bank note or a pinch of salt.
Ounce/aʊns/
Symbol: ozUnit System: Imperial
### What is the Ounce?
The oz is a unit of mass (acceptable for use as weight on Earth) and also is part of the imperial system of units. It has the price oz.
Not to be perplexed with a liquid ounce (fl oz; volume) or an ounce-force (force), the ounce is the smallest of the 3 denominations the weight provided in the royal system. There space 16 ounces in a pound and 14 pounds in a stone.
The strict surname for this unit is the avoirdupois ounce and in SI / metric state it is indistinguishable to around 28.3g.
See more: What Can You Say About A Really Terrible Mummy Joke ? Mummy Worksheet (One
Ounces are provided to show the load of fabrics in Asia, the UK and North America. For example, 16 oz denim. The number describes weight that the cloth in ounces.
The oz was no longer seen as a legitimate unit of measure up after the year 2000 in the UK. However, the is still offered informally and also is likewise used together the measure up for part sizes in restaurants in the UK.
## Conversion Tables for Grams (g) to Ounces (oz)
0g - 240g
0g = 0oz10g = 0.35oz20g = 0.71oz30g = 1.06oz40g = 1.41oz50g = 1.76oz60g = 2.12oz70g = 2.47oz80g = 2.82oz90g = 3.17oz100g = 3.53oz110g = 3.88oz120g = 4.23oz130g = 4.59oz140g = 4.94oz150g = 5.29oz160g = 5.64oz170g = 6oz180g = 6.35oz190g = 6.7oz200g = 7.05oz210g = 7.41oz220g = 7.76oz230g = 8.11oz240g = 8.47oz
250g - 490g
250g = 8.82oz260g = 9.17oz270g = 9.52oz280g = 9.88oz290g = 10.23oz300g = 10.58oz310g = 10.93oz320g = 11.29oz330g = 11.64oz340g = 11.99oz350g = 12.35oz360g = 12.7oz370g = 13.05oz380g = 13.4oz390g = 13.76oz400g = 14.11oz410g = 14.46oz420g = 14.82oz430g = 15.17oz440g = 15.52oz450g = 15.87oz460g = 16.23oz470g = 16.58oz480g = 16.93oz490g = 17.28oz
500g - 740g
500g = 17.64oz510g = 17.99oz520g = 18.34oz530g = 18.7oz540g = 19.05oz550g = 19.4oz560g = 19.75oz570g = 20.11oz580g = 20.46oz590g = 20.81oz600g = 21.16oz610g = 21.52oz620g = 21.87oz630g = 22.22oz640g = 22.58oz650g = 22.93oz660g = 23.28oz670g = 23.63oz680g = 23.99oz690g = 24.34oz700g = 24.69oz710g = 25.04oz720g = 25.4oz730g = 25.75oz740g = 26.1oz
750g - 990g
750g = 26.46oz760g = 26.81oz770g = 27.16oz780g = 27.51oz790g = 27.87oz800g = 28.22oz810g = 28.57oz820g = 28.92oz830g = 29.28oz840g = 29.63oz850g = 29.98oz860g = 30.34oz870g = 30.69oz880g = 31.04oz890g = 31.39oz900g = 31.75oz910g = 32.1oz920g = 32.45oz930g = 32.8oz940g = 33.16oz950g = 33.51oz960g = 33.86oz970g = 34.22oz980g = 34.57oz990g = 34.92oz
We"re just a group of passionates do a tool to aid students, engineers and also the human populace navigate the crazy people of unit conversion through a little bit of ease!
Something missing? want to contribute? get in touch!
Convert
Popular Conversions
| 1,357
| 3,596
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.03125
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.903725
|
http://www.intmath.com/blog/mathematics/roller-door-problem-7665
| 1,503,141,312,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-34/segments/1502886105341.69/warc/CC-MAIN-20170819105009-20170819125009-00673.warc.gz
| 595,607,823
| 20,248
|
# Roller Door problem
By Murray Bourne, 28 Nov 2012
In an earlier article, I discussed the Length of an Archimedean Spiral. We saw we can find the length of such a spiral using integration, once we know the equation of the spiral and the beginning and end points. The equation for an Archimedean Spiral is in polar form, and is determined by an angle θ, the amount of turn of the spiral.
Roller door.
[Image source]
Brent, an engineer with a roller door company, asked an interesting question in the comments for that article. He posed a "real-life" issue he faced.
His problem is the reverse. He knows length of the spiral already (it's the height of the roller door, plus the bit that's wrapped around the drum wheel), but he needs to know the number of turns of the drum for the given height of the door and radius of the drum.
This is not such a straightforward problem.
I suggested he could produce a table for already known door heights and number of turns. and then use interpolation to find the required number of turns (at least approximately).
Brent also turned to PhysicsForums to ask for help. He wrote:
Hi,
I'm an engineer designing a spring system for a garage roller door. I need to know the number of turns of the door for all the size combinations.
I've found this page which gives a good equation for finding the length if you know the number of turns, starting radius and gap between spirals:
$L=\int_a^b\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}d\theta$
The equation of the spiral is r = x + yθ, with x = starting radius, y = gap/(), and to find L we're taking the integral from a = 0 to b = 2πn (where n = turns).
When you know n, this is straightfoward, and even I could work that out. But it's been a decade since I've done anything like this, so I was wondering if anyone could help me solve for n in this integral:
$L=\int_0^{(2\pi){n}}\sqrt{(x+y\theta)^2+y^2}d\theta$
My way of solving this is to find L for n=1,2,3,4,5 etc, graph it in Excel and use "find trendline" to get an equation. Any help appreciated, thanks.
The response there was not very helpful.
Brent provided his Excel spreadsheet so I could see what he was doing.
## Specific Example
Brent's roller door has a barrel radius of 127.5 cm and the thickness of the door slats (called the "curtain") is 4.0 cm. This means the radius will increase by 4.0 cm for each turn of the barrel.
So x = 127.5 and y = 4/(2π)=0.6366 in this case.
The following daigram (not to scale) indicates our situation:
Applying the integration formula above, we have the length of the curtain wrapped around the barrel for 1 turn:
$L=\int_0^{2\pi}\sqrt{(127.5+0.6366\theta)^2+0.6366^2}d\theta=813.68$
Similarly, for 2 turns (that is, 0 to 4π), we have:
$L=\int_0^{4\pi}\sqrt{(127.5+0.6366\theta)^2+0.6366^2}d\theta=1654.4$
Continuing on for 3 up to 6 turns, we obtain the following values:
Turns Length
0
0
1
814
2
1654
3
2516
4
3406
5
4320
6
5259
Next, we use Excel to plot these values and to draw a trendline through them. We choose "polynomial of degree 2" because a consideration of the integral and an inspection of the Excel graph (when extrapolating beyond 6 turns) shows it is parabolic.
Excel can tell us the equation for the graph produced, and it is included on the chart.
We need to translate the "y" and "x" variables on Excel's chart for our own situation. The vertical scale is the length of the curtain, and the horizontal scale is the number of turns. Also, the "0.1667" value is a rounding error, which we can discard.
So in our case, we would write the Length of the curtain (L) for the number of turns, n as:
$L=12.5n^2+801.4n$
Solving for n (using the quadratic formula) gives us 2 solutions. The first one is:
$n=0.04\sqrt{(50.0L+6.4224\times10^5)}-32.056$
The second solution gives a negative result for n, so we discard it.
We now check our work (by comparing a known value).
If the length is 3406 cm, then the number of turns given by our new formula is:
$n=0.04\sqrt{(50.0\times3406+6.4224\times10^5)}-32.056=4.0004$
This is certainly close to the 4 turns we expected.
So Brent could now use this formula to find the number of turns for any roller door's curtain length given a barrel radius of 127.5 cm and a curtain thickness of 4 cm.
## Linear Solution
The Excel graph for the range of turns 0 to 6 is very nearly linear. Using the "Linear" trendline option in Excel gives the model for the Length given the number of turns as:
L = 876.5n - 62.4
Solving for n gives:
$n=\frac{L+62.4}{876.5}$
Applying this simpler formula to some of our known values gives:
For length L = 814, number of turns, n = 0.99989. (expected 1)
For length L = 1654, number of turns, n = 1.9582. (expected 2)
For length L = 5529, number of turns, n = 6.0712. (expected 6)
So for lengths of the roller door curtain less than about 10 m, this simpler linear model gives acceptable results.
## Other solutions
Neil provided another solution in the comments.
He proposed:
$n=\frac{OL}{\pi(ID+OD)}$
where
n = number of turns
OL = overall length
ID = inner diameter of the spiral
OD = outer diameter of the spiral
However, Brent pointed out this can't be used because he doesn't know the outer diameter until he knows the number of turns.
Jacob then weighed in with another proposal, which involved taking an average of the inner and outer radius and then applying the circumference of a circle to that average.
Jacob suggests if we don't know the number of turns, we can use this formula to find it:
$N=\frac{y-ID+\sqrt{(ID-y)^2+\frac{4yL}{\pi}}}{2y}$
Applying that formula to our example above, with y = 4 and ID = 127.5×2 = 255 and our known case where L = 3406 gives:
$N=\frac{4-255+\sqrt{(255-4)^2+\frac{4(4)(3406)}{\pi}}}{2(4)}=4.0571$
This is very close to the 4 turns we expected.
Applying the formula for another known case, L = 5259, we obtain 6.08 turns, nicely close to the expected value of 6.
## Conclusion
Brent has a choice of formulas to use, all of which give acceptable results. The linear model is the easiest.
Parabolic model: $n=0.04\sqrt{(50.0L+6.4224\times10^5)}-32.056$
Linear model: $n=\frac{L+62.4}{876.5}$
The model suggested by Neil and refined by Jacob:
$N=\frac{4-255+\sqrt{(255-4)^2+\frac{4(4)(L)}{\pi}}}{2(4)}$
Here is a table that shows the values from the given formulas.
Length
(via calculus)
Turns
(original)
Turns
(parabolic model)
Turns
(linear model)
Turns
(Neil/Jacob model)
0 0 0 0 0
814 1 1.0001 0.9999 1.0158
1654 2 2.0014 1.9582 2.0318
2516 3 2.9992 2.9417 3.0431
3406 4 4.0004 3.9571 4.0571
4320 5 5.0005 4.9999 5.069
5259 6 6.0006 6.0712 6.0802
### 9 Comments on “Roller Door problem”
1. Thomas A Buckley says:
Two different simple non calculus methods yielded identical results, which differ greatly from the table used to derive the concluding solution above. So calculus people, recheck!
For example you say starting radius x = 127.5 cm. Taking center of door thickness for length, allowing inner shortening to be balanced by outer lengthening when rolled,
effective starting radius = 127.5 + 4/2 = 129.5 cm!
.....................................Length cm....Length cm
1.........................2Pi(127.5+2)=814........880
2............2Pi(127.5+6)+814=1652.......1917
3.....2Pi(127.5+10)+1652=2516.......3111
4.....2Pi(127.5+14)+2516=3405.......4462
5.....2Pi(127.5+18)+3405=4320.......5970
6.....2Pi(127.5+22)+4320=5259.......7636
After the 1st turn each diameter increases by 4 cm.
OK, so my method is NOT exact, but it should not be so different from the results obtained by calculus.
2nd Method.
Number of Turns n
= [SqRt( Lt/pi + r^2)-r]/t
L = Door length, t = door thickness, r = barrel radius.
CSA of door = Lt
This area must fit in the space between concentric circles.
Taking R as radius when door is wound on barrel, then
Lt = piR^2 - pir^2 . . ie both areas are the same.
R = SqRt( Lt/pi + r^2) Eq1
Thickness of door on barrel = R-r
Number of turns n required
= (R-r) /t . . now substitute RHS of Eq1 for R
= [SqRt( Lt/pi + r^2)-r]/t
Note R disappears, and providing the desired variables,
door Length and thickness, and barrel radius, yields the required number of Turns. Also fractional turns are valid, eg 4.5 turns means 4.5 turns. This equation gives the same results as my 1st method table above.
Still cannot get html codes in my equations, sorry Murray.
2. Thomas A Buckley says:
Jacob's formula returns 5.2 turns. The 12.8 turns result is in error because barrel DIAMETER is 127.5 x2=255, not 127.5/2!
My equation returns 5.15 turns. That is good agreement
for different approximate methods, casting further doubt that the variables for door length of 4462cm requires only 4 turns as the earlier table shows.
3. Alan says:
Brent's best bet is to get Mathcad! A numerical solution is then obtained almost trivially in a couple of lines.
4. Murray says:
@Thomas - Thank you! The errors were due to my not dividing the curtain width 4 by 2pi.
I have re-written and fixed that whole part, plus added a simpler linear model.
Your approximations using the average radius certainly make sense for this problem, and align very closely with the (corrected) values via calculus.
I must have been half asleep when looking at Jacob's formula. That part has been corrected in the article, too!
@Alan - Good suggestion, but Mathcad is probably overkill for this problem...
5. brent says:
Alan - I needed to obtain a formula because I needed to do this calculation about 1000 times: for every size and width of door.
Thanks.
6. Calla says:
Hello! I am using this problem in a mathematical exploration assignment for school. I understand most of the solution, except for the line, 'x = 127.5 and y = 4/(2?)=0.6366 in this case'. Why do we divide the width of the curtain (4 cm) by 2pi? What do x and y refer to in this line?
Any help would be greatly appreciated.
7. Calla says:
I am confused by one aspect of this article: the author divides 4 (the distance between each arm) by 2pi in order to get b, 0.6366. This follows Brent's assertion that in the equation r=x + y pheta, y=gap/2pi.
But in a previous article ("Length of an Archimedean Spiral") it was stated that,
Total angle turned by spiral in radians = 2?b
Distance between each arm (or 'gap') = 2?b divided by the number of turns
Thus, how can b be derived from dividing 2pi by the gap? I would think that:
b = (gap x number of turns) divided by 2?
based on the previous article. Could anyone help in explaining this to me? It would be greatly appreciated.
8. Eric Blaise says:
I am actually glad I do not have to worry about this. I am pretty sure I did a similar problem in geometry but cannot use it today for the life of me. Most garage doors continue rolling until they encounter resistance and then stop. either way, I am pretty sure there is an application for this somewhere.
Eric
9. brent says:
Eric, the number of turns is needed to calculate how many turns the springs need to sized for. You can't do any spring calculations without having a good way to calculate turns of the door.
### Comment Preview
HTML: You can use simple tags like <b>, <a href="...">, etc.
To enter math, you can can either:
1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$and$$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)
NOTE: You can't mix both types of math entry in your comment.
| 3,251
| 11,597
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 14, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625
| 4
|
CC-MAIN-2017-34
|
latest
|
en
| 0.948876
|
https://niniloos.com/complex-one-digit-addition-worksheets/
| 1,620,773,094,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243990419.12/warc/CC-MAIN-20210511214444-20210512004444-00140.warc.gz
| 453,354,065
| 8,426
|
# Complex one digit addition worksheets information
» » Complex one digit addition worksheets information
Your Complex one digit addition worksheets images are ready. Complex one digit addition worksheets are a topic that is being searched for and liked by netizens today. You can Get the Complex one digit addition worksheets files here. Find and Download all free images.
If you’re searching for complex one digit addition worksheets images information connected with to the complex one digit addition worksheets interest, you have visit the right site. Our website always gives you hints for seeing the maximum quality video and image content, please kindly surf and find more enlightening video articles and graphics that match your interests.
Complex One Digit Addition Worksheets. Two Plus One Digit Addition. Number of columns 1 columns 2 columns 3 columns 4 columns 5 columns 6 columns. 12092020 One Digit Addition Worksheets Are your children facing difficulties to understand the one-digit addition. 31082018 These worksheets are available as part of the Add 3-digit and 1-digit numbers - crossing 10 lesson pack.
Christmas Single Digit Addition Worksheet Have Fun Teaching Addition Worksheets Christmas Math Worksheets Christmas Addition From pinterest.com
26032020 Single-digit addition and subtraction worksheets that will be helpful for the beginning students. Follow the links for Spaceship Math Subtraction worksheets timed subtraction tests multiple digit subtraction worksheets simple borrowing and regrouping worksheets and math worksheets with mixed addition and subtraction problems. A set of differentiated activity sheets for children to practise complex partitioning of 3-digit numbersIf your students need a bit more support with three-digit numbers try these Base Ten Blocks Worksheets too for a lovely visual resource. If you are a student of a college a teachers aide then you will be required to perform all kinds of calculations on your worksheets in order to make it more comprehensible. Missing addends sums to 10 20. Some of the worksheets displayed are Grade 1 addition work Grade 2 addition work Mental math adding 3 digit and 1 digit numbers Adding a 2 digit number and a 1 digit number missing addend 3 digit subtraction Addition 3 digit subtraction 1 Mixed operations work adding and subtracting three.
### Let your kids practise addition and subtraction through these vertical equation worksheets.
Source: pinterest.com
26032020 Single-digit addition and subtraction worksheets that will be helpful for the beginning students. Why not to try our exercise so your child learns more through practice. Number line worksheets help the kids understand how the numbers will be added. Addition - 1 Digit Browse to find free 1 digit addition worksheets created by teachers. 24052020 Workbook addition worksheet is one of the most essential items for any business of any type and organization.
Source: pinterest.com
Free 1 Digit Addition Math Worksheet. This math worksheet was created on 2019-01-09 and has been viewed 42 times this week and 344 times this month. 31082018 These worksheets are available as part of the Add 3-digit and 1-digit numbers - crossing 10 lesson pack. Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade 5th Grade 6th Grade. Each of these addition worksheets provide exercises for progressively more complex types of addition problems with multiple digit addends as well as problems with more than two addends.
Grade 1 addition worksheets. The two most common varieties for both simple and complex worksheets is the Block-By and Line-By worksheets. Let your kids practise addition and subtraction through these vertical equation worksheets. Number of columns 1 columns 2 columns 3 columns 4 columns 5 columns 6 columns. On the other hand the Line-By worksheet uses a column format in which you write down one digit and then another digit that follows.
Source: prodigygame.com
Isnt that fun adding. 12092020 One Digit Addition Worksheets Are your children facing difficulties to understand the one-digit addition. This math worksheet was created on 2019-01-09 and has been viewed 42 times this week and 344 times this month. The worksheets are perfect for Kindergarten and year1 students. Why not to try our exercise so your child learns more through practice.
Addition 1 Digit Addition 1 Digit - Displaying top 8 worksheets found for this concept. A set of differentiated activity sheets for children to practise complex partitioning of 3-digit numbersIf your students need a bit more support with three-digit numbers try these Base Ten Blocks Worksheets too for a lovely visual resource. There is no login or membership required on this free teacher resource website. The worksheets are editable giving you the option to make any changes you feel necessary with the knowledge you have about your class and school setting. The numbers for each addend may be individually varied to generate different problem sets for these worksheets.
Source: pinterest.com
A worksheet with addition problems set out in a column format. Some of the worksheets displayed are Grade 1 addition work Grade 2 addition work Mental math adding 3 digit and 1 digit numbers Adding a 2 digit number and a 1 digit number missing addend 3 digit subtraction Addition 3 digit subtraction 1 Mixed operations work adding and subtracting three. Showing top 8 worksheets in the category - Adding 3 Digit Number And 1 Digit Number. This math worksheet was created on 2019-01-09 and has been viewed 42 times this week and 344 times this month. The worksheets are perfect for Kindergarten and year1 students.
Source: pinterest.com
Addition - 1 Digit Grade - 1st Grade - Kindergarten Grade -PreK - TK Math Numbers - 1 to 10. Let your kids practise addition and subtraction through these vertical equation worksheets. Single Digit Addition Worksheets Vertical Format - 2 Addends These single digit addition worksheets are configured for 2 addends in a vertical problem format. Addition - 1 Digit Browse to find free 1 digit addition worksheets created by teachers. Isnt that fun adding.
This is the main page for the subtraction worksheets. Number of columns 1 columns 2 columns 3 columns 4 columns 5 columns 6 columns. Single-digit addition worksheets include addition drills addition statement addition wheel circle the sum and more. Access some of these worksheets for free. The worksheets are editable giving you the option to make any changes you feel necessary with the knowledge you have about your class and school setting.
Source: mathworksheets4kids.com
Free 1 Digit Addition Math Worksheet. Follow the links for Spaceship Math Subtraction worksheets timed subtraction tests multiple digit subtraction worksheets simple borrowing and regrouping worksheets and math worksheets with mixed addition and subtraction problems. The worksheets are editable giving you the option to make any changes you feel necessary with the knowledge you have about your class and school setting. Each of these addition worksheets provide exercises for progressively more complex types of addition problems with multiple digit addends as well as problems with more than two addends. Two Plus One Digit Addition.
Source: pinterest.com
Free 1 Digit Addition Math Worksheet. Addition - 1 Digit Browse to find free 1 digit addition worksheets created by teachers. Single-digit addition worksheets include addition drills addition statement addition wheel circle the sum and more. If you are a student of a college a teachers aide then you will be required to perform all kinds of calculations on your worksheets in order to make it more comprehensible. Addition 1 Digit Addition 1 Digit - Displaying top 8 worksheets found for this concept.
This is the main page for the subtraction worksheets. If you are working with more complex calculation then. Simply find a free printable worksheet and click to download. Number of columns 1 columns 2 columns 3 columns 4 columns 5 columns 6 columns. Number line worksheets help the kids understand how the numbers will be added.
Source: pinterest.com
The numbers for each addend may be individually varied to generate different problem sets for these worksheets. Check out my freebies to get a taste for the quality of resources available. A set of differentiated activity activities sheets for children to practise complex partitioning of 3-digit numbers. Number line worksheets help the kids understand how the numbers will be added. This math worksheet was created on 2019-01-09 and has been viewed 42 times this week and 344 times this month.
Source: pinterest.com
11 is 2 22 is 4. The worksheets are perfect for Kindergarten and year1 students. 11 is 2 22 is 4. 26032020 Single-digit addition and subtraction worksheets that will be helpful for the beginning students. In addition to this there are also mixed problems worksheets given in end for practice.
Source: worksheetsplanet.com
A set of differentiated activity sheets for children to practise complex partitioning of 3-digit numbersIf your students need a bit more support with three-digit numbers try these Base Ten Blocks Worksheets too for a lovely visual resource. 31082018 These worksheets are available as part of the Add 3-digit and 1-digit numbers - crossing 10 lesson pack. Addition - 1 Digit Browse to find free 1 digit addition worksheets created by teachers. 24052020 Workbook addition worksheet is one of the most essential items for any business of any type and organization. Some of the worksheets displayed are Grade 1 addition work Grade 2 addition work Mental math adding 3 digit and 1 digit numbers Adding a 2 digit number and a 1 digit number missing addend 3 digit subtraction Addition 3 digit subtraction 1 Mixed operations work adding and subtracting three.
Source: pinterest.com
Adding numbers up to 10 or 30 by counting pictured objects. Isnt that fun adding. A worksheet with addition problems set out in a column format. Addition - 1 Digit Grade - 1st Grade - Kindergarten Grade -PreK - TK Math Numbers - 1 to 10. A set of differentiated activity sheets for children to practise complex partitioning of 3-digit numbersIf your students need a bit more support with three-digit numbers try these Base Ten Blocks Worksheets too for a lovely visual resource.
This site is an open community for users to submit their favorite wallpapers on the internet, all images or pictures in this website are for personal wallpaper use only, it is stricly prohibited to use this wallpaper for commercial purposes, if you are the author and find this image is shared without your permission, please kindly raise a DMCA report to Us.
If you find this site helpful, please support us by sharing this posts to your preference social media accounts like Facebook, Instagram and so on or you can also save this blog page with the title complex one digit addition worksheets by using Ctrl + D for devices a laptop with a Windows operating system or Command + D for laptops with an Apple operating system. If you use a smartphone, you can also use the drawer menu of the browser you are using. Whether it’s a Windows, Mac, iOS or Android operating system, you will still be able to bookmark this website.
| 2,225
| 11,248
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.625
| 3
|
CC-MAIN-2021-21
|
latest
|
en
| 0.833003
|
https://www.bartleby.com/solution-answer/chapter-35-problem-54e-single-variable-calculus-concepts-and-contexts-enhanced-edition-4th-edition/9781337687805/959d820e-5563-11e9-8385-02ee952b546e
| 1,632,539,630,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057589.14/warc/CC-MAIN-20210925021713-20210925051713-00379.warc.gz
| 679,712,024
| 66,028
|
The function f ( x ) = x + e x is one to one.
Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
Solutions
Chapter 3.5, Problem 54E
(a)
To determine
To show: The function f(x)=x+ex is one to one.
Expert Solution
Explanation of Solution
Given:
The function is f(x)=x+ex.
Proof:
Obtain the derivative of the function f(x)=x+ex.
Differentiate implicitly with respect to x,
f(x)=ddx(x+ex)=ddx(x)+ddx(ex)=1+ex
Here, the value ex is always positive, then f(x)=1+ex is positive.
Note that, if f(x) is positive then the function f(x) is increasing function.
Thus, f(x)=x+ex is increasing function, every element of the codomain has at most one preimage.
Therefore, the function f(x) is one to one.
Hence the required result is proved.
(b)
To determine
To find: The values of f−1(1).
Expert Solution
The value f1(1)=0.
Explanation of Solution
Calculation:
Obtain the value of f1(1).
Let x=f1(1). That is, compute f(x)=1.
Substitute x=0 in f(x)=x+ex,
f(0)=0+e0=0+1=1
Thus, the value is f(0)=1.
Since f(x) is one to one,f1(1)=0.
Therefore, the value f1(1)=0.
(c)
To determine
To find: The value of the f−1(1) (use formula from ex 77 (a))
Expert Solution
The value (f1)(1)=12.
Explanation of Solution
Formula used:
The derivative of inverse function (f1)(x)=1f(f1(x)).
Calculation:
Given function is f(x)=x+ex.
Substitute 1 for x in (f1)(x)=1f(f1(x)),
(f1)(1)=1f(f1(1))
From part (a) and (b), the derivative is f(x)=1+ex and the value of f1(1)=0.
Substitute 0 for x in f(x)=1+ex,
f(0)=1+e0=1+1=2
Thus, the value is f(0)=2.
Substitute f1(1)=0 in (f1)(1),
(f1)(1)=1f(0)=12 [Qf(0)=2]
Therefore, the value (f1)(1)=12.
Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!
| 651
| 2,033
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5625
| 5
|
CC-MAIN-2021-39
|
latest
|
en
| 0.719904
|
https://gitee.com/scruel/ML-AndrewNg-Notes/blob/master/week3.md
| 1,611,634,529,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-04/segments/1610704795033.65/warc/CC-MAIN-20210126011645-20210126041645-00323.warc.gz
| 355,319,164
| 36,411
|
## scruel / ML-AndrewNg-Notes .gitee-modal { width: 500px !important; }
Create your Gitee Account
Explore and code with more than 6 million developers，Free private repositories ！：）
This repository doesn't specify license. Without author's permission, this code is only for learning and cannot be used for other purposes.
week3.md 22.18 KB
scruel authored 2019-05-25 13:51 . images path in htmls
[TOC]
# 6 逻辑回归(Logistic Regression)
## 6.1 分类(Classification)
• 垃圾邮件判断
• 金融欺诈判断
• 肿瘤诊断
$h_\theta(x) \geqslant 0.5$ ，预测为 $y = 1$，即正向类；
$h_\theta(x) \lt 0.5$ ，预测为 $y = 0$，即负向类。
## 6.2 假设函数表示(Hypothesis Representation)
sigmoid 函数（如下图）是逻辑函数的特殊情况，其公式为 $g\left( z \right)=\frac{1}{1+{{e}^{-z}}}$。
\begin{align*}& h_\theta(x) = P(y=1 | x ; \theta) = 1 - P(y=0 | x ; \theta) \newline & P(y = 0 | x;\theta) + P(y = 1 | x ; \theta) = 1\end{align*}
## 6.3 决策边界(Decision Boundary)
\begin{align*}& h_\theta(x) \geq 0.5 \rightarrow y = 1 \newline& h_\theta(x) < 0.5 \rightarrow y = 0 \newline\end{align*}
${h_\theta}\left( x \right)=g\left( {\theta_0}+{\theta_1}{x_1}+{\theta_{2}}{x_{2}}+{\theta_{3}}x_{1}^{2}+{\theta_{4}}x_{2}^{2} \right)$
## 6.4 代价函数(Cost Function)
\begin{align*}& J(\theta) = \dfrac{1}{m} \sum_{i=1}^m \mathrm{Cost}(h_\theta(x^{(i)}),y^{(i)}) \newline & \mathrm{Cost}(h_\theta(x),y) = -\log(h_\theta(x)) ; & \text{if y = 1} \newline & \mathrm{Cost}(h_\theta(x),y) = -\log(1-h_\theta(x)) ; & \text{if y = 0}\end{align*}
## 6.5 简化的成本函数和梯度下降(Simplified Cost Function and Gradient Descent)
$J(\theta) = - \frac{1}{m} \displaystyle \sum_{i=1}^m [y^{(i)}\log (h_\theta (x^{(i)})) + (1 - y^{(i)})\log (1 - h_\theta(x^{(i)}))]$
$h = g(X\theta)$，$J(\theta) = \frac{1}{m} \cdot \left(-y^{T}\log(h)-(1-y)^{T}\log(1-h)\right)$
\begin{align*} & \text{repeat until convergence:} ; \lbrace \newline ; &{{\theta }{j}}:={{\theta }{j}}-\alpha \frac{\partial }{\partial {{\theta }_{j}}}J\left( {\theta} \right) \newline \rbrace \end{align*}
\begin{align*}& \text{repeat until convergence:} ; \lbrace \newline ; & \theta_j := \theta_j - \alpha \frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)} ; & \text{for j := 0,1...n}\newline \rbrace\end{align*}
$J(\theta) = - \frac{1}{m} \displaystyle \sum_{i=1}^m [y^{(i)}\log (h_\theta (x^{(i)})) + (1 - y^{(i)})\log (1 - h_\theta(x^{(i)}))]$
$f(\theta)={{y}^{(i)}}\log \left( \frac{1}{1+{{e}^{-z}}} \right)+\left( 1-{{y}^{(i)}} \right)\log \left( 1-\frac{1}{1+{{e}^{-z}}} \right)$ $=-{{y}^{(i)}}\log \left( 1+{{e}^{-z}} \right)-\left( 1-{{y}^{(i)}} \right)\log \left( 1+{{e}^{z}} \right)$
$\frac{\partial z}{\partial {\theta_{j}}}=\frac{\partial }{\partial {\theta_{j}}}\left( \theta^Tx^{(i)} \right)=x^{(i)}_j$
$\frac{\partial }{\partial {\theta_{j}}}f\left( \theta \right)=\frac{\partial }{\partial {\theta_{j}}}[-{{y}^{(i)}}\log \left( 1+{{e}^{-z}} \right)-\left( 1-{{y}^{(i)}} \right)\log \left( 1+{{e}^{z}} \right)]$
$=-{{y}^{(i)}}\frac{\frac{\partial }{\partial {\theta_{j}}}\left(-z \right) e^{-z}}{1+e^{-z}}-\left( 1-{{y}^{(i)}} \right)\frac{\frac{\partial }{\partial {\theta_{j}}}\left(z \right){e^{z}}}{1+e^{z}}$
$=-{{y}^{(i)}}\frac{-x^{(i)}_je^{-z}}{1+e^{-z}}-\left( 1-{{y}^{(i)}} \right)\frac{x^{(i)}_j}{1+e^{-z}}$ $=\left({{y}^{(i)}}\frac{e^{-z}}{1+e^{-z}}-\left( 1-{{y}^{(i)}} \right)\frac{1}{1+e^{-z}}\right)x^{(i)}_j$ $=\left({{y}^{(i)}}\frac{e^{-z}}{1+e^{-z}}-\left( 1-{{y}^{(i)}} \right)\frac{1}{1+e^{-z}}\right)x^{(i)}j$ $=\left(\frac{{{y}^{(i)}}(e^{-z}+1)-1}{1+e^{-z}}\right)x^{(i)}j$ $={({{y}^{(i)}}-\frac{1}{1+{{e}^{-z}}})x_j^{(i)}}$ $={\left({{y}^{(i)}}-{h\theta}\left( {{x}^{(i)}} \right)\right)x_j^{(i)}}$ $=-{\left({h\theta}\left( {{x}^{(i)}} \right)-{{y}^{(i)}}\right)x_j^{(i)}}$
$\frac{\partial }{\partial {\theta_{j}}}J(\theta) = -\frac{1}{m}\sum\limits_{i=1}^{m}{\frac{\partial }{\partial {\theta_{j}}}f(\theta)}=\frac{1}{m} \sum\limits_{i=1}^{m} (h_\theta(x^{(i)}) - y^{(i)}) \cdot x_j^{(i)}$
## 6.6 进阶优化(Advanced Optimization)
• 梯度下降法(Gradient Descent)
• 共轭梯度算法(Conjugate gradient)
• 牛顿法和拟牛顿法(Newton's method & Quasi-Newton Methods)
• DFP算法
• 局部优化法(BFGS)
• 有限内存局部优化法(L-BFGS)
• 拉格朗日乘数法(Lagrange multiplier)
Octave/Matlab 中对这类高级算法做了封装，易于调用。
1. 创建一个函数以返回代价函数及其偏导数：
function [jVal, gradient] = costFunction(theta)
% code to compute J(theta)
jVal=(theta(1)-5)^2+(theta(2)-5)^2;
% code to compute derivative of J(theta)
gradient=zeros(2,1);
gradient(1)=2*(theta(1)-5);
gradient(2)=2*(theta(2)-5);
end
1. costFunction 函数及所需参数传入最优化函数 fminunc，以求解最优化问题：
options = optimset('GradObj', 'on', 'MaxIter', 100);
initialTheta = zeros(2,1);
[optTheta, functionVal, exitFlag] = fminunc(@costFunction, initialTheta, options);
'GradObj', 'on': 启用梯度目标参数（则需要将梯度传入算法）
'MaxIter', 100: 最大迭代次数为 100 次
@xxx: Octave/Matlab 中的函数指针
optTheta: 最优化得到的参数向量
functionVal: 引用函数最后一次的返回值
exitFlag: 标记代价函数是否收敛
1. 返回结果
optTheta =
5
5
functionVal = 0
exitFlag = 1
## 6.7 多类别分类: 一对多(Multiclass Classification: One-vs-all)
$h_\theta^{\left( i \right)}\left( x \right)$: 输出 $y=i$（属于第 $i$ 个分类）的可能性
$k$: 类别总数，如上图 $k=3$。
# 7 正则化(Regularization)
## 7.1 过拟合问题(The Problem of Overfitting)
• 欠拟合(Underfitting)
无法很好的拟合训练集中的数据，预测值和实际值的误差很大，这类情况被称为欠拟合。拟合模型比较简单（特征选少了）时易出现这类情况。类似于，你上课不好好听，啥都不会，下课也差不多啥都不会。
• 优良的拟合(Just right)
不论是训练集数据还是不在训练集中的预测数据，都能给出较为正确的结果。类似于，学霸学神！
• 过拟合(Overfitting)
能很好甚至完美拟合训练集中的数据，即 $J(\theta) \to 0$，但是对于不在训练集中的新数据，预测值和实际值的误差会很大，泛化能力弱，这类情况被称为过拟合。拟合模型过于复杂（特征选多了）时易出现这类情况。类似于，你上课跟着老师做题都会都听懂了，下课遇到新题就懵了不会拓展。
• 偏差(bias)
指模型的预测值与真实值的偏离程度。偏差越大，预测值偏离真实值越厉害。偏差低意味着能较好地反应训练集中的数据情况。
• 方差(Variance)
指模型预测值的离散程度或者变化范围。方差越大，数据的分布越分散，函数波动越大，泛化能力越差。方差低意味着拟合曲线的稳定性高，波动小。
• 减少特征的数量
• 手动选取需保留的特征
• 使用模型选择算法来选取合适的特征(如 PCA 算法)
• 减少特征的方式易丢失有用的特征信息
• 正则化(Regularization)
• 可保留所有参数（许多有用的特征都能轻微影响结果）
• 减少/惩罚各参数大小(magnitude)，以减轻各参数对模型的影响程度
• 当有很多参数对于模型只有轻微影响时，正则化方法的表现很好
## 7.2 代价函数(Cost Function)
$min_\theta\ \dfrac{1}{2m}\sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})^2 + 1000\cdot\theta_3^2 + 1000\cdot\theta_4^2$
$J\left( \theta \right)=\frac{1}{2m}[\sum\limits_{i=1}^{m}{{{({h_\theta}({{x}^{(i)}})-{{y}^{(i)}})}^{2}}+\lambda \sum\limits_{j=1}^{n}{\theta_{j}^{2}}]}$
$\lambda$: 正则化参数(Regularization Parameter)，$\lambda > 0$
$\sum\limits_{j=1}^{n}$: 不惩罚基础参数 $\theta_0$
$\lambda \sum\limits_{j=1}^{n}{\theta_{j}^{2}}$: 正则化项
$\lambda$ 正则化参数类似于学习速率，也需要我们自行对其选择一个合适的值。
• 过大
• 导致模型欠拟合(假设可能会变成近乎 $x = \theta_0$ 的直线 )
• 无法正常去过拟问题
• 梯度下降可能无法收敛
• 过小
• 无法避免过拟合（等于没有）
## 7.3 线性回归正则化(Regularized Linear Regression)
\begin{align*} & \text{Repeat}\ \lbrace \newline & \ \ \ \ \theta_0 := \theta_0 - \alpha\ \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)} \newline & \ \ \ \ \theta_j := \theta_j - \alpha\ \left[ \left( \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)} \right) + \frac{\lambda}{m}\theta_j \right], \ \ \ j \in \lbrace 1,2...n\rbrace\newline & \rbrace \end{align*}
$\theta_j := \theta_j(1 - \alpha\frac{\lambda}{m}) - \alpha\frac{1}{m}\sum_{i=1}^m(h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)}$
$\frac{\lambda}{m}\theta_j$: 正则化项
\begin{align*}& \theta = \left( X^TX + \lambda \cdot L \right)^{-1} X^Ty \newline& \text{where}\ \ L = \begin{bmatrix} 0 & & & & \newline & 1 & & & \newline & & 1 & & \newline & & & \ddots & \newline & & & & 1 \newline\end{bmatrix}\end{align*}
$\lambda\cdot L$: 正则化项
$L$: 第一行第一列为 $0$ 的 $n+1$ 维单位矩阵
Matlab/Octave 代码：
>> L = eye(5)
>> L(1,1) = 0
L =
0 0 0 0 0
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1
## 7.4 逻辑回归正则化(Regularized Logistic Regression)
$J(\theta) = - \frac{1}{m} \sum_{i=1}^m \large[ y^{(i)}\ \log (h_\theta (x^{(i)})) + (1 - y^{(i)})\ \log (1 - h_\theta(x^{(i)}))\large] + \frac{\lambda}{2m}\sum_{j=1}^n \theta_j^2$
\begin{align*} & \text{Repeat}\ \lbrace \newline & \ \ \ \ \theta_0 := \theta_0 - \alpha\ \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_0^{(i)} \newline & \ \ \ \ \theta_j := \theta_j - \alpha\ \left[ \left( \frac{1}{m}\ \sum_{i=1}^m (h_\theta(x^{(i)}) - y^{(i)})x_j^{(i)} \right) + \frac{\lambda}{m}\theta_j \right], \ \ \ j \in \lbrace 1,2...n\rbrace\newline & \rbrace \end{align*}
[^2]: week2 - 4.6
### Comment ( 0 )
Sign in for post a comment
Matlab
1
https://gitee.com/scruel/ML-AndrewNg-Notes.git
git@gitee.com:scruel/ML-AndrewNg-Notes.git
scruel
ML-AndrewNg-Notes
ML-AndrewNg-Notes
master
| 4,002
| 8,282
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.453125
| 3
|
CC-MAIN-2021-04
|
latest
|
en
| 0.295877
|
http://bactra.org/weblog/1178.html
| 1,632,769,309,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780058467.95/warc/CC-MAIN-20210927181724-20210927211724-00578.warc.gz
| 6,113,867
| 4,043
|
## March 26, 2021
### Regression, Thermostats, Causal Inference: Some Finger Exercises
$\newcommand{\Expect}[1]{\mathbb{E}\left[ #1 \right]} \newcommand{\Prob}[1]{\mathbb{P}\left( #1 \right)} \newcommand{\Cov}[1]{\mathrm{Cov}\left[ #1 \right]} \newcommand{\Var}[1]{\mathrm{Var}\left[ #1 \right]}$
Attention conservation notice: An 800-word, literally academic exercise about an issue in causal inference. Its point is familiar to those in the field, and deservedly obscure to everyone else. Also, too cutesy and pleased with itself by at least half.
I wrote the first version of this for the class where we do causal inference long enough ago that I actually don't remember when --- 2011? 2013? (In retrospect I had probably read Milton Friedman's thermostat analogy but didn't consciously remember it at the time.) Posted now because I've gone over the point with two different people in the last month.
The temperature outside $(X)$ is a direct cause of the temperature inside my house $(Y)$. But every morning I measure the temperature, and adjust my heating/cooling system $(C)$ to try to maintain a constant temperature $y_0$. For simplicity, we'll say that all the relations are linear, so $\begin{eqnarray} X & \sim & \mathrm{whatever}\\ C|X & \leftarrow & a+bX + \epsilon_1\\ Y|X,C & \leftarrow & X-C + \epsilon_2 \end{eqnarray}$ where $\epsilon_1$ and $\epsilon_2$ are exogenous, independent, mean-zero noise terms. We can think of $\epsilon_1$ as a combination of my sloppiness in measuring the temperature and in tuning the heating/cooling system; $\epsilon_2$ is sheer fluctuations.
Exercise: Draw the DAG.
To ensure that the expectation of $Y$ remains at $y_0$, no matter the external temperature, we need $\begin{eqnarray} y_0 & = & \Expect{Y|X=x}\\ & = & \Expect{X - a + bX + \epsilon_1 + \epsilon_2|X=x}\\ & = & (1-b)x -a \end{eqnarray}$ Since this must hold for all $x$, we need $b=1, a=-y_0$.
What follows from this?
• Internal temperature $Y$ is uncorrelated with external temperature $X$: $\begin{eqnarray} \Cov{X,Y} & = & \Expect{XY} - \Expect{X}\Expect{Y}\\ & = & \Expect{X\Expect{Y|X}} - \Expect{X}\Expect{Y}\\ & = & \Expect{X}y_0 - \Expect{X}y_0 = 0 \end{eqnarray}$ The internal temperature will fluctuate around the set-point $y_0$, but those fluctuations will not correlate with the external temperature.
• Internal temperature $Y$ is correlated with the control signal $C$ only through my sloppiness: $\begin{eqnarray} \Cov{C,Y} & = & \Expect{CY} - \Expect{C}\Expect{Y}\\ & = & \Expect{(-y_0 + X + \epsilon_1)(X+y_0-X-\epsilon_1+\epsilon_2)} - (\Expect{X}-y_0)y_0\\ & = & -y_0^2 - \Expect{\epsilon^2} + \Expect{X}y_0 -\Expect{X \epsilon_1} + \Expect{X\epsilon_2} + \Expect{\epsilon_1 \epsilon_2} - \Expect{X}y_0 + y_0^2\\ & = & -\Var{\epsilon_1} \end{eqnarray}$ since all the cross-expectations are zero, and $\Expect{\epsilon_1}=0$.
• The control signal $C$ is correlated with the external temperature: $\begin{eqnarray} \Cov{C,X} & = & \Expect{CX} - \Expect{C}\Expect{X}\\ & = & \Expect{(-y_0 + X+\epsilon_1)X} + (-y_0 +\Expect{X})\Expect{X}\\ & = & \Expect{X^2} - \left(\Expect{X}\right)^2\\ & = & \Var{X} \end{eqnarray}$
• A linear regression of $Y$ on $X$ and $C$ will consistently recover the correct coefficients, namely $+1$ and $-1$. To see this, recall (e.g., from here) that the OLS estimates will tend towards the coefficients of the optimal linear predictor. Those coefficients, in turn, are the solution to $\beta = {\left[ \begin{array}{cc} \Var{X} & \Cov{C,X}\\ \Cov{X,C} & \Var{C} \end{array}\right]}^{-1} \left[ \begin{array}{c} \Cov{Y,X}\\ \Cov{Y,C} \end{array}\right]$ Plugging in our previous results, $\beta = {\left[ \begin{array}{cc} \Var{X} & \Var{X}\\ \Var{X} & \Var{X}+\Var{\epsilon_1} \end{array}\right]}^{-1} \left[ \begin{array}{c} 0\\ -\Var{\epsilon} \end{array}\right]$ After some character-building algebra, you can confirm that the covariance matrix is invertible as long as $\Var{\epsilon_1} > 0$, and then, as promised $\beta = (1,-1)$.
Exercise: Build your character by doing the algebra.
So, as long as control isn't perfect, the naive statistician (or experienced econometrician...) who just does a kitchen-sink regression will actually get the relationship between $Y$, $X$ and $C$ right, concluding that external temperature and the climate control have equal and opposite effects on internal temperature. Sure, there will be sampling noise, but with enough data they'll approach the truth.
Exercise: What do you get if you regress $C$ on $X$ and $Y$?
I have implicitly assumed that I know the exact linear relationship between $X$ and $Y$, since I used that in deriving how the control signal should respond to $X$. If I mis-calibrate the control signal, say if $C = -y_0 +0.999X + \epsilon_1$, then there is not an exact cancellation and everything works as usual.
Exercise: Suppose that instead of measuring the external temperature $X$ directly, I can only measure yesterday's temperature $U$, again with noise. Supposing there is a linear relationship between $U$ and $X$, replicate this analysis. Does it matter if $U$ is the parent of $X$ or vice versa?
Exercise: "Feedback is a mechanism for persistently violating faithfulness"; discuss.
Exercise: "The greatest skill seems like clumsiness" (Laozi); discuss.
Posted at March 26, 2021 09:08 | permanent link
| 1,588
| 5,349
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.40625
| 3
|
CC-MAIN-2021-39
|
latest
|
en
| 0.828418
|
https://simplywall.st/stocks/au/telecom/asx-sda/speedcast-international-shares/news/speedcast-international-limited-asxsda-delivered-a-weaker-roe-than-its-industry/
| 1,596,773,366,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439737152.0/warc/CC-MAIN-20200807025719-20200807055719-00580.warc.gz
| 406,535,583
| 15,449
|
# SpeedCast International Limited (ASX:SDA) Delivered A Weaker ROE Than Its Industry
Want to participate in a short research study? Help shape the future of investing tools and you could win a \$250 gift card!
One of the best investments we can make is in our own knowledge and skill set. With that in mind, this article will work through how we can use Return On Equity (ROE) to better understand a business. By way of learning-by-doing, we’ll look at ROE to gain a better understanding of SpeedCast International Limited (ASX:SDA).
Our data shows SpeedCast International has a return on equity of 4.0% for the last year. That means that for every A\$1 worth of shareholders’ equity, it generated A\$0.040 in profit.
### How Do I Calculate ROE?
The formula for ROE is:
Return on Equity = Net Profit ÷ Shareholders’ Equity
Or for SpeedCast International:
4.0% = 11.268 ÷ US\$284m (Based on the trailing twelve months to June 2018.)
Most know that net profit is the total earnings after all expenses, but the concept of shareholders’ equity is a little more complicated. It is all the money paid into the company from shareholders, plus any earnings retained. The easiest way to calculate shareholders’ equity is to subtract the company’s total liabilities from the total assets.
### What Does ROE Signify?
ROE looks at the amount a company earns relative to the money it has kept within the business. The ‘return’ is the amount earned after tax over the last twelve months. That means that the higher the ROE, the more profitable the company is. So, all else equal, investors should like a high ROE. That means it can be interesting to compare the ROE of different companies.
### Does SpeedCast International Have A Good ROE?
By comparing a company’s ROE with its industry average, we can get a quick measure of how good it is. The limitation of this approach is that some companies are quite different from others, even within the same industry classification. As shown in the graphic below, SpeedCast International has a lower ROE than the average (14%) in the Telecom industry classification.
That’s not what we like to see. We prefer it when the ROE of a company is above the industry average, but it’s not the be-all and end-all if it is lower. Nonetheless, it might be wise to check if insiders have been selling.
### Why You Should Consider Debt When Looking At ROE
Companies usually need to invest money to grow their profits. That cash can come from issuing shares, retained earnings, or debt. In the first two cases, the ROE will capture this use of capital to grow. In the latter case, the debt required for growth will boost returns, but will not impact the shareholders’ equity. In this manner the use of debt will boost ROE, even though the core economics of the business stay the same.
### Combining SpeedCast International’s Debt And Its 4.0% Return On Equity
SpeedCast International clearly uses a significant amount debt to boost returns, as it has a debt to equity ratio of 1.65. The combination of a rather low ROE and significant use of debt is not particularly appealing. Debt increases risk and reduces options for the company in the future, so you generally want to see some good returns from using it.
### The Bottom Line On ROE
Return on equity is a useful indicator of the ability of a business to generate profits and return them to shareholders. In my book the highest quality companies have high return on equity, despite low debt. If two companies have the same ROE, then I would generally prefer the one with less debt.
Having said that, while ROE is a useful indicator of business quality, you’ll have to look at a whole range of factors to determine the right price to buy a stock. Profit growth rates, versus the expectations reflected in the price of the stock, are a particularly important to consider. So I think it may be worth checking this free report on analyst forecasts for the company.
If you would prefer check out another company — one with potentially superior financials — then do not miss this free list of interesting companies, that have HIGH return on equity and low debt.
To help readers see past the short term volatility of the financial market, we aim to bring you a long-term focused research analysis purely driven by fundamental data. Note that our analysis does not factor in the latest price-sensitive company announcements.
The author is an independent contributor and at the time of publication had no position in the stocks mentioned. For errors that warrant correction please contact the editor at editorial-team@simplywallst.com.
| 981
| 4,620
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.53125
| 3
|
CC-MAIN-2020-34
|
latest
|
en
| 0.9284
|
https://numycyvygojeqo.douglasishere.com/write-an-equation-for-a-fourth-degree-polynomial-with-no-real-zeros-44422sj.html
| 1,620,265,262,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243988724.75/warc/CC-MAIN-20210505234449-20210506024449-00288.warc.gz
| 453,620,796
| 5,305
|
# Write an equation for a fourth degree polynomial with no real zeros
He also developed the earliest techniques of the infinitesimal calculus; Archimedes credits Eudoxus with inventing a principle eventually called the Axiom of Archimedes: For example, there are 25 primes less than and 23 luckies less than Hyperreals include all the reals in the technical sense that they form an ordered field containing the reals as a subfield and also contain infinitely many other numbers that are either infinitely large numbers whose absolute value is greater than any positive real number or infinitely small numbers whose absolute value is less than any positive real number.
He studied optical illusions and was first to explain psychologically why the Moon appears to be larger when near the horizon. Students will visit at least one relevant site, exhibit or museum as a course requirement.
CAD topics include software commands and drawing strategies for 2-D and 3-D CAD work, plans, sections, elevations, and details, information management, assembly of drawings and scales. Only fragments survive but it apparently used axiomatic-based proofs similar to Euclid's and contains many of the same theorems.
Panini has been called "one of the most innovative people in the whole development of knowledge;" his grammar "one of the greatest monuments of human intelligence. I'm preparing a page to consider that question.
Al-Kindi, called The Arab Philosopher, can not be considered among the greatest of mathematicians, but was one of the most influential general scientists between Aristotle and da Vinci. He developed the Stomachion puzzle and solved a difficult enumeration problem involving it ; other famous gems include The Cattle-Problem.
Copernicus, Bruno, Galileo and Kepler lived 14 centuries after Ptolemy. Students must either complete a paper or poster at the conclusion of their research internship. He worked in plane and spherical trigonometry, and with cubic equations.
This may happen because we have data from different sources we want to combine, or because we organize the code with variables that are easy to read, and then want to combine the variables. Their zeros are at zero, negative squares of two, and positive squares of two. Among several novel achievements in astronomy, he used observations of lunar eclipse to deduce relative longitude, estimated Earth's radius most accurately, believed the Earth rotated on its axis and may have accepted heliocentrism as a possibility.
Well, if you subtract nine from both sides, you get x-squared is equal to negative nine. Long denied legitimacy in mathematics, negative numbers are nowhere to be found in the writings of the Babylonians, Greeks, or other ancient cultures. Put another way, whereas the set of all rationals is countable, the irrationals form an uncountable set and therefore represent a larger kind of infinity.
Represent verbal statements of multiplicative comparisons as multiplication equations. To avoid confusion, 0, 1, 2, 3, And we're getting ahead of our story: In these cases, you have to employ smoothing techniques, either implicitly by using a multipoint derivative formula, or explicitly by smoothing the data yourself, or taking the derivative of a function that has been fit to the data in the neighborhood you are interested in.
We consider those in the next section. So, those are our zeros. In algebra, he solved various equations including 2nd-order Diophantine, quartic, Brouncker's and Pell's equations. They surface for the first time in in bookkeeping records seventh-century India and in a chapter of a work by the Hindu astronomer Brahmagupta.
Although Euler and Newton may have been the most important mathematicians, and Gauss, Weierstrass and Riemann the greatest theorem provers, it is widely accepted that Archimedes was the greatest genius who ever lived.
Unlike our system, with ten digits separate from the alphabet, the 27 Greek number symbols were the same as their alphabet's letters; this might have hindered the development of "syncopated" notation.
Factoring a Polynomial A quadratic factor with no real zeros is said to be prime or irreducible over the reals. x2 + 1 = (x – i)(x + i) is irreducible over the reals (and therefore over the rationals). x2 – 2 = is irreducible over the rationals but reducible over the reals.
- [Voiceover] So, we have a fifth-degree polynomial here, p of x, and we're asked to do several things. First, find the real roots. And let's sort of remind ourselves what roots are. SOLUTION: Write a fourth degree polynomial equation with integer coefficients that has two irrational roots and two imaginary roots.
Algebra -> Polynomials-and-rational-expressions -> SOLUTION: Write a fourth degree polynomial equation with integer coefficients that. Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation.
This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. See explanation Note that if a polynomial has Real coefficients, then any non-Real Complex zeros occur in Complex conjugate pairs.
So to construct a quartic with no Real zeros, start with two pairs of Complex conjugate numbers. May 04, · Best Answer: No (if you mean "exactly", rather than "at least", 3 complex zeros) A polynomial has exactly as many zeros (including both real and complex) as its degree.
So all 4th degree polynomials have 4 douglasishere.com: Resolved.
Write an equation for a fourth degree polynomial with no real zeros
Rated 0/5 based on 48 review
SOLUTION: Write the equation of the polynomial function with the given zeros: 3,4,-2
| 1,204
| 5,748
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.84375
| 3
|
CC-MAIN-2021-21
|
latest
|
en
| 0.95505
|
https://byjus.com/question-answer/in-the-given-figure-ab-de-are-bc-if-ab-9-cm-de-3-cm/
| 1,723,275,379,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722640790444.57/warc/CC-MAIN-20240810061945-20240810091945-00453.warc.gz
| 121,313,130
| 25,056
|
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question
# In the given figure, AB & DE are ⊥ BC, if AB = 9 cm, DE = 3 cm and AC = 24 cm, find AD
A
8 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
16 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
E
32 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
## The correct option is B 16 cm
Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Pythogoras Theorem
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program
| 242
| 752
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2024-33
|
latest
|
en
| 0.813177
|
http://library.kiwix.org/wikipedia_en_computer_novid_2018-10/A/Closed_timelike_curve.html
| 1,558,676,291,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232257514.68/warc/CC-MAIN-20190524044320-20190524070320-00274.warc.gz
| 115,277,143
| 10,539
|
# Closed timelike curve
In mathematical physics, a closed timelike curve (CTC) is a world line in a Lorentzian manifold, of a material particle in spacetime that is "closed", returning to its starting point. This possibility was first discovered by Willem Jacob van Stockum in 1937[1] and later confirmed by Kurt Gödel in 1949,[2] who discovered a solution to the equations of general relativity (GR) allowing CTCs known as the Gödel metric; and since then other GR solutions containing CTCs have been found, such as the Tipler cylinder and traversable wormholes. If CTCs exist, their existence would seem to imply at least the theoretical possibility of time travel backwards in time, raising the spectre of the grandfather paradox, although the Novikov self-consistency principle seems to show that such paradoxes could be avoided. Some physicists speculate that the CTCs which appear in certain GR solutions might be ruled out by a future theory of quantum gravity which would replace GR, an idea which Stephen Hawking has labeled the chronology protection conjecture. Others note that if every closed timelike curve in a given space-time passes through an event horizon, a property which can be called chronological censorship, then that space-time with event horizons excised would still be causally well behaved and an observer might not be able to detect the causal violation.[3]
## Light cones
When discussing the evolution of a system in general relativity, or more specifically Minkowski space, physicists often refer to a "light cone". A light cone represents any possible future evolution of an object given its current state, or every possible location given its current location. An object's possible future locations are limited by the speed that the object can move, which is at best the speed of light. For instance, an object located at position p at time t0 can only move to locations within p + c(t1 t0) by time t1.
This is commonly represented on a graph with physical locations along the horizontal axis and time running vertically, with units of for time and ct for space. Light cones in this representation appear as lines at 45 degrees centered on the object, as light travels at per . On such a diagram, every possible future location of the object lies within the cone. Additionally, every space location has a future time, implying that an object may stay at any location in space indefinitely.
Any single point on such a diagram is known as an event. Separate events are considered to be timelike if they are separated across the time axis, or spacelike if they differ along the space axis. If the object were in free fall, it would travel up the t-axis; if it accelerates, it moves across the x axis as well. The actual path an object takes through spacetime, as opposed to the ones it could take, is known as the worldline. Another definition is that the light cone represents all possible worldlines.
In "simple" examples of spacetime metrics the light cone is directed forward in time. This corresponds to the common case that an object cannot be in two places at once, or alternately that it cannot move instantly to another location. In these spacetimes, the worldlines of physical objects are, by definition, timelike. However this orientation is only true of "locally flat" spacetimes. In curved spacetimes the light cone will be "tilted" along the spacetime's geodesic. For instance, while moving in the vicinity of a star, the star's gravity will "pull" on the object, affecting its worldline, so its possible future positions lie closer to the star. This appears as a slightly tilted lightcone on the corresponding spacetime diagram. An object in free fall in this circumstance continues to move along its local axis, but to an external observer it appears it is accelerating in space as well—a common situation if the object is in orbit, for instance.
In extreme examples, in spacetimes with suitably high-curvature metrics, the light cone can be tilted beyond 45 degrees. That means there are potential "future" positions, from the object's frame of reference, that are spacelike separated to observers in an external rest frame. From this outside viewpoint, the object can move instantaneously through space. In these situations the object would have to move, since its present spatial location would not be in its own future light cone. Additionally, with enough of a tilt, there are event locations that lie in the "past" as seen from the outside. With a suitable movement of what appears to it its own space axis, the object appears to travel through time as seen externally.
A closed timelike curve can be created if a series of such light cones are set up so as to loop back on themselves, so it would be possible for an object to move around this loop and return to the same place and time that it started. An object in such an orbit would repeatedly return to the same point in spacetime if it stays in free fall. Returning to the original spacetime location would be only one possibility; the object's future light cone would include spacetime points both forwards and backwards in time, and so it should be possible for the object to engage in time travel under these conditions.
## General relativity
CTCs appear in locally unobjectionable exact solutions to the Einstein field equation of general relativity, including some of the most important solutions. These include:
• the Misner space (which is Minkowski space orbifolded by a discrete boost)
• the Kerr vacuum (which models a rotating uncharged black hole)
• the interior of a rotating BTZ black hole
• the van Stockum dust (which models a cylindrically symmetric configuration of dust)
• the Gödel lambdadust (which models a dust with a carefully chosen cosmological constant term)
• the Tipler cylinder (a cylindrically symmetric metric with CTCs)
• Bonnor-Steadman solutions describing laboratory situations such as two spinning balls
• J. Richard Gott has proposed a mechanism for creating CTCs using cosmic strings.
Some of these examples are, like the Tipler cylinder, rather artificial, but the exterior part of the Kerr solution is thought to be in some sense generic, so it is rather unnerving to learn that its interior contains CTCs. Most physicists feel that CTCs in such solutions are artifacts.
## Consequences
One feature of a CTC is that it opens the possibility of a worldline which is not connected to earlier times, and so the existence of events that cannot be traced to an earlier cause. Ordinarily, causality demands that each event in spacetime is preceded by its cause in every rest frame. This principle is critical in determinism, which in the language of general relativity states complete knowledge of the universe on a spacelike Cauchy surface can be used to calculate the complete state of the rest of spacetime. However, in a CTC, causality breaks down, because an event can be "simultaneous" with its cause—in some sense an event may be able to cause itself. It is impossible to determine based only on knowledge of the past whether or not something exists in the CTC that can interfere with other objects in spacetime. A CTC therefore results in a Cauchy horizon, and a region of spacetime that cannot be predicted from perfect knowledge of some past time.
No CTC can be continuously deformed as a CTC to a point (that is, a CTC and a point are not timelike homotopic), as the manifold would not be causally well behaved at that point. The topological feature which prevents the CTC from being deformed to a point is known as a timelike topological feature.
Existence of CTCs places restrictions on physically allowable states of matter-energy fields in the universe. Propagating a field configuration along the family of closed timelike worldlines must eventually result in the state that is identical to the original one. This has been explored by some scientists as a possible approach towards disproving the existence of CTCs.
Existence of CTCs implies also equivalence of quantum and classical computation (both in PSPACE).[4]
## Contractible versus noncontractible
There are two classes of CTCs. We have CTCs contractible to a point (if we no longer insist it has to be future-directed timelike everywhere), and we have CTCs which are not contractible. For the latter, we can always go to the universal covering space, and reestablish causality. For the former, such a procedure is not possible. No closed timelike curve is contractible to a point by a timelike homotopy among timelike curves, as that point would not be causally well behaved.[3]
## Cauchy horizon
The chronology violating set is the set of points through which CTCs pass. The boundary of this set is the Cauchy horizon. The Cauchy horizon is generated by closed null geodesics. Associated with each closed null geodesic is a redshift factor describing the rescaling of the rate of change of the affine parameter around a loop. Because of this redshift factor, the affine parameter terminates at a finite value after infinitely many revolutions because the geometric series converges.
## Notes
1. Stockum, W. J. van (1937). "The gravitational field of a distribution of particles rotating around an axis of symmetry.". Proc. Roy. Soc. Edinburgh. 57.
2. Stephen Hawking, My Brief History, chapter 11
3. H. Monroe (2008). "Are Causality Violations Undesirable?". Foundations of Physics. 38 (11): 1065&ndash, 1069. arXiv:gr-qc/0609054. Bibcode:2008FoPh...38.1065M. doi:10.1007/s10701-008-9254-9.
4. Watrous, John; Aaronson, Scott (2009). "Closed timelike curves make quantum and classical computing equivalent". Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences. 465 (2102): 631. arXiv:0808.2669. Bibcode:2009RSPSA.465..631A. doi:10.1098/rspa.2008.0350.
## References
• S. Carroll (2004). Spacetime and Geometry. Addison Wesley. ISBN 0-8053-8732-3.
• Kurt Gödel (1949). "An Example of a New Type of Cosmological Solution of Einstein's Field Equations of Gravitation". Rev. Mod. Phys. 21 (3): 447. Bibcode:1949RvMP...21..447G. doi:10.1103/RevModPhys.21.447.
• W. Bonnor; B.R. Steadman (2005). "Exact solutions of the Einstein-Maxwell equations with closed timelike curves". Gen. Rel. Grav. 37 (11): 1833. Bibcode:2005GReGr..37.1833B. doi:10.1007/s10714-005-0163-3.
• Joe Haldeman (2008). The Accidental Time Machine.
| 2,371
| 10,389
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2019-22
|
latest
|
en
| 0.962076
|
https://link.springer.com/article/10.1007/s11858-018-01022-8?error=cookies_not_supported&code=2672c428-32bd-46f2-a5a9-6095c0d6576a
| 1,621,015,617,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991428.43/warc/CC-MAIN-20210514152803-20210514182803-00177.warc.gz
| 374,598,262
| 34,197
|
# Student understanding of linear combinations of eigenvectors
## Abstract
To contribute to the sparse educational research on student understanding of eigenspace, we investigated how students reason about linear combinations of eigenvectors. We present results from student reasoning on two written multiple-choice questions with open-ended justifications involving linear combinations of eigenvectors in which the resultant vector is or is not an eigenvector of the matrix. We detail seven themes that analysis of our data revealed regarding student responses. These themes include: determining if a linear combination of eigenvectors satisfies the equation $$A\varvec{x}=\lambda \varvec{x}$$; reasoning about a linear combination of eigenvectors belonging to a set of eigenvectors; conflating scalars in a linear combination with eigenvalues; thinking eigenvectors must be linearly independent; and reasoning about the number of eigenspace dimensions for a matrix. In the discussion, we explore how themes sometimes cut across questions and how looking across questions gives insight into individuals’ conceptions of eigenspace. Implications for teaching and future research are also offered.
This is a preview of subscription content, access via your institution.
## Notes
1. 1.
“An eigenvector of an $$nxn$$ matrix $$A$$ is a nonzero vector $$\varvec{x}$$ such that $$A\varvec{x}=\lambda \varvec{x}$$ for some scalar $$\lambda$$. A scalar $$\lambda$$ is called an eigenvalue of $$A$$ if there is a nontrivial solution $$\varvec{x}$$ of $$A\varvec{x}=\lambda \varvec{x}$$; such a $$\varvec{x}$$ is called an eigenvector corresponding to$$\lambda$$” (Lay et al., 2016, p. 269). The eigenspace of $$A$$ corresponding to $$\lambda$$ is “the set of all solutions of $$A\varvec{x}=\lambda \varvec{x}$$, where $$\lambda$$ is an eigenvalue of $$A.$$ [It c]onsists of the zero vector and all eigenvectors corresponding to $$\lambda$$” (p. A9).
2. 2.
This paper builds from and is an extension of a conference presentation given at the 2018 Research in Undergraduate Mathematics Education Conference (Wawro, Watson, & Zandieh, 2018).
3. 3.
“A genetic decomposition is a hypothetical model that describes the mental structures and mechanisms that a student might need to construct in order to learn a specific mathematical concept” (Arnon et al., 2014, p. 27).
4. 4.
See Tall (2004) for a more detailed discussion of the Three Worlds of Mathematics (geometric/embodied, symbolic, formal) used as part of the framing in Thomas and Stewart (2011).
5. 5.
“Process conception” and “object conception” are constructs from APOS Theory; see, for example, Dubinsky and McDonald (2001) for more information.
6. 6.
The main MCE version prompts students to justify their answer to the multiple-choice stem by selecting all pre-made justification statements that support their choice (Watson et al., 2017; Zandieh, Plaxco, Wawro, Rasmussen, Milbourne, & Czeranko 2015).
## References
1. Arnon, I., Cottrill, J., Dubinsky, E., Oktac, A., Roa Fuentes, S., Trigueros, M., et al. (2014). APOS Theory. A framework for research and curriculum development in mathematics education. Nueva York: Springer.
2. Artigue, M., Batanero, C., & Kent, P. (2007). Mathematics thinking and learning at post-secondary level. In F. K. Lester Jr.. (Ed.), Second handbook of research on mathematics teaching and learning (pp. 1011–1050). Reston: National Council of Teachers of Mathematics.
3. Beltrán-Meneu, M. J., Murillo-Arcila, M., & Albarracín, L. (2016). Emphasizing visualization and physical applications in the study of eigenvectors and eigenvalues. Teaching Mathematics and its Applications: An International Journal of the IMA, 36(3), 123–135.
4. Blumer, H. (1969). Symbolic interactionism: Perspectives and method. Englewood Cliffs: Prentice-Hall.
5. Bouhjar, K., Andrews-Larson, C., Haider, M., & Zandieh, M. (2018). Examining Students’ Procedural and Conceptual Understanding of Eigenvectors and Eigenvalues in the Context of Inquiry-Oriented Instruction. In S. Stewart, C. Andrews-Larson, A. Berman, & M. Zandieh (Eds.), Challenges In Teaching Linear Algebra Challenges and Strategies in Teaching Linear Algebra (pp. 193–216). Cham: Springer.
6. Çağlayan, G. (2015). Making sense of eigenvalue-eigenvector relationships: Math majors’ linear algebra–geometry connections in a dynamic environment. Journal of Mathematical Behavior, 40, 131–153.
7. Cobb, P., & Yackel, E. (1996). Constructivist, emergent, and sociocultural perspectives in the context of developmental research. Educational Psychologist, 31, 175–190.
8. Dorier, J.-L. (Ed.). (2000). On the teaching of linear algebra. Dordrecht: Kluwer Academic.
9. Dubinsky, E., & McDonald, M. (2001). APOS: A constructivist theory of learning. In D. Holton (Ed.), The teaching and learning of mathematics at university level: An ICMI study (pp. 275–282). Dordrecht: Kluwer Academic.
10. Glaser, B., & Strauss, A. (1967). The discovery of grounded theory: Strategies for qualitative research. Chicago: Aldine Publishing Company.
11. Gol Tabaghi, S., & Sinclair, N. (2013). Using dynamic geometry software to explore eigenvectors: The emergence of dynamic-synthetic-geometric thinking. Technology, Knowledge and Learning, 18(3), 149–164.
12. Harel, G. (2000). Three principles of learning and teaching mathematics. In J.-L. Dorier (Ed.), On the teaching of linear algebra. Dordrecht: Kluwer Academic.
13. Henderson, F., Rasmussen, C., Sweeney, G., Wawro, M., & Zandieh, M. (2010). Symbol sense in linear algebra: A start toward eigen theory. Proceedings of the 13th Annual Conference on Research in Undergraduate Mathematics Education, Raleigh, NC. Retrieved from http://sigmaa.maa.org/rume/crume2010. Accessed 1 Sept 2015.
14. Hiebert, J., & Lafevre, P. (1986). Conceptual and procedural knowledge in mathematics: An introductory analysis. In J. Hiebert (Ed.), Conceptual and procedural knowledge: The case of mathematics (pp. 1–27). Hillsdale: Lawrence Erlbaum Associates.
15. Hillel, J. (2000). Modes of description and the problem of representation in linear algebra. In J.-L. Dorier (Ed.), On the teaching of linear algebra (pp. 191–207). Dordrecht: Kluwer.
16. Larson, C., & Zandieh, M. (2013). Three interpretations of the matrix equation Ax = b. For the Learning of Mathematics, 33(2), 11–17.
17. Lay, D., Lay, S., & McDonald, J. (2016). Linear algebra and its applications (5th edn.). Essex: Pearson Education.
18. Lesh, R., Hoover, M., Hole, B., Kelly, A., & Post, T. (2000). Principles for developing thought-revealing activities for students and teachers. In A. Kelly & R. Lesh (Eds.), Handbook of research design in mathematics and science education (pp. 591–646). New Jersey: Lawrence Erlbaum.
19. Miles, M. B., Huberman, A. M., & Saldaña, J. (2014). Fundamentals of qualitative data analysis. In Qualitative Data Analysis: A Methods Sourcebook (3rd edn.). Thousand Oaks: Sage Publications Inc.
20. Nyman, M. A., Lapp, D. A., St. John, D., & Berry, J. S. (2010). Those do what? Connecting eigenvectors and eigenvalues to the rest of linear algebra: Using visual enhancements to help students connect eigenvectors to the rest of linear algebra. International Journal for Technology in Mathematics Education, 17(1), 35–41.
21. Plaxco, D., Zandieh, M., & Wawro, M. (2018). Stretch directions and stretch factors: A sequence intended to support guided reinvention of eigenvector and eigenvalue. In S. Stewart, C. Andrews-Larson, A. Berman & M. Zandieh (Eds.), Challenges In Teaching Linear Algebra. ICME-13 Monographs (pp. 175–192). Cham: Springer.
22. Rasmussen, C., & Wawro, M. (2017). Post-calculus research in undergraduate mathematics education. In J. Cai (Ed.), The compendium for research in mathematics education (pp. 551–579). Reston: National Council of Teachers of Mathematics.
23. Salgado, H., & Trigueros, M. (2015). Teaching eigenvalues and eigenvectors using models and APOS Theory. The Journal of Mathematical Behavior, 39, 100–120.
24. Saxe, G. B. (2002). Children’s developing mathematics in collective practices: A framework for analysis. Journal of the Learning Sciences, 11, 275–300.
25. Tall, D. O. (2004). Building theories: The three worlds of mathematics. For the Learning of Mathematics, 24(1), 29–32.
26. Thomas, M. O. J., & Stewart, S. (2011). Eigenvalues and eigenvectors: Embodied, symbolic and formal thinking. Mathematics Education Research Journal, 23(3), 275–296.
27. von Glasersfeld, E. (1995). Radical constructivism: A way of knowing and learning. Bristol: Falmer Press.
28. Watson, K., Wawro, M., Zandieh, M., & Kerrigan, S. (2017). Knowledge about student understanding of eigentheory: Information gained from multiple choice extended assessment. In A. Weinberg, C. Rasmussen, J. Rabin, M. Wawro, and S. Brown (Eds.), Proceedings of the 20th annual conference on research in undergraduate mathematics education (pp. 311–325), San Diego, CA: SIGMAA on RUME.
29. Wawro, M., Watson, K., & Zandieh, M. (2018). Student understanding of linear combinations of eigenvectors. In A. Weinberg, C. Rasmussen, J. Rabin, M. Wawro, and S. Brown (Eds.), Proceedings of the 21st annual conference on research in undergraduate mathematics education (pp. 1372–1378), San Diego, CA: SIGMAA on RUME.
30. Wawro, M., Zandieh, M., Rasmussen, C., & Andrews-Larson, C. (2013). Inquiry oriented linear algebra: Course materials. Retrieved from http://iola.math.vt.edu. Accessed 1 Sept 2015.
31. Wawro, M., Zandieh, M., & Watson, K. (2018). Delineating aspects of understanding eigentheory through assessment development. In V. Durand-Guerrier, R. Hochmuth, S. Goodchild, & N.M. Hogstad (Eds.), Proceedings of INDRUM 2018 Second Conference of the International Network for Didactic Research in University Mathematics (pp. 275–284), Kristiansand: University of Agder and INDRUM.
32. Zandieh, M., Adiredja, A., & Knapp, J. (2018). Exploring Everyday Examples to ExplainBasis from Eight German Male Graduate STEM Students. (Manuscript submitted for publication).
33. Zandieh, M., & Andrews-Larson, C. (2018). Solving linear system: Reconstructing unknowns to interpret row reduced matrices. (Manuscript submitted for publication).
34. Zandieh, M., Plaxco, D., Wawro, M., Rasmussen, C., Milbourne, H., & Czeranko, K. (2015). Extending multiple choice format to document student thinking. In T. Fukawa-Connelly, N. Infante, K. Keene, and M. Zandieh (Eds.), Proceedings of the 18th annual conference on research in undergraduate mathematics education (pp. 1094–1100), Pittsburgh, PA: SIGMAA on RUME.
35. Zandieh, M., Wawro, M., & Rasmussen, C. (2017). An example of inquiry in linear algebra: The roles of symbolizing and brokering. PRIMUS, 27(1), 96–124.
## Acknowledgements
This material is based upon work supported by the National Science Foundation under Grant Number DUE-1452889. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the authors and do not necessarily reflect the views of the National Science Foundation.
## Author information
Authors
### Corresponding author
Correspondence to Megan Wawro.
## Rights and permissions
Reprints and Permissions
Wawro, M., Watson, K. & Zandieh, M. Student understanding of linear combinations of eigenvectors. ZDM Mathematics Education 51, 1111–1123 (2019). https://doi.org/10.1007/s11858-018-01022-8
| 3,076
| 11,397
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125
| 4
|
CC-MAIN-2021-21
|
latest
|
en
| 0.843838
|
https://nursingacademics.com/2021/06/23/the-following-data-give-the-selling-price-square-footage-number-of/
| 1,679,590,333,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296945182.12/warc/CC-MAIN-20230323163125-20230323193125-00757.warc.gz
| 484,749,868
| 15,844
|
# The following data give the selling price, square footage, number of
Problem 4-22
The following data give the selling price, square footage, number of bedrooms and age of houses that have sold in a neighborhood in the past 6 months. Develop 3 regression models to predict the selling price based upon each of the other factors individually. Which of these is best?
Selling price Square footageBedrooms Average years
84,000 1,670 2 30
79,000 1,339 2 25
91,500 1,712 3 30
120,0001,840 3 40
127,5002,300 3 18
132,5002,234 3 30
145,0002,311 3 19
164,0002,377 3 7
155,0002,736 4 10
168,0002,500 3 1
172,5002,500 4 3
174,0002,479 3 3
175,0002,400 3 1
177,5003,124 4 0
184,0002,500 3 2
195,5004,062 4 10
195,0002,854 3 3
Problem 4-23
Use the data in problem 4-22 and develop a regression model to predict selling price based on the square footage and number of bedrooms. Use this to predict the selling price of a 2,000 square foot house with three bedrooms. Compare this model with the models in problem 4-22. Should the number of bedrooms be included in the model? Why or why not?
1. State the linear equation.
2. Explain the overall statistical significance of the model.
3. Explain the statistical significance for each independent variable in the model.
4. Interpret the adjusted R2.
5. Is this a good predictive equation(s)? Which variables should be excluded, if any, and why? Explain.
Problem 4-24
Use the data in problem 4-22 and develop a regression model to predict selling price based on the square footage, number of bedrooms, and age. Use this to predict the selling price of a 10 year old, 2,000 square foot house with three bedrooms.
Problem 4-30
In 2012 the total payroll for the New York Yankees was almost \$200 million, while the total payroll for the Oakland Athletics (a team known for using baseball analytics or sabermetrics) was about \$55 million, less than 1/3 of the Yankees payroll. In the following table, you will see the payrolls in millions and the total number of victories for the baseball teams in the American League in the 2012 season. Develop a regression model to predict the total number of victories based on the payroll. Use the model to predict the number of victories for a team with a payroll of \$79 million. Based on the results of the computer output, discuss the relationship between payroll and victories.
Team Payroll in millions Number of victories
Baltimore Orioles 81.4 93
Boston Red Sox 173.2 69
Chicago White Sox 96.9 85
Cleveland Indians 78.4 68
Detroit Tigers 132.3 88
Kansas City Royals 60.9 72
Los Angeles Angels 154.5 89
Minnesota Twins 94.1 66
NY Yankees 198.0 95
Oakland Athletics 55.4 94
Seattle Mariners 82.0 75
Tampa Bay Rays 64.2 90
Texas Rangers 120.5 93
Toronto Blue Jays 75.5 73
| 938
| 4,783
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625
| 4
|
CC-MAIN-2023-14
|
latest
|
en
| 0.613932
|
https://engineering.stackexchange.com/questions/58731/calculating-heat-loss-of-water-in-a-tube
| 1,716,422,550,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058575.96/warc/CC-MAIN-20240522224707-20240523014707-00578.warc.gz
| 198,717,812
| 39,638
|
# Calculating heat loss of water in a tube
I am seeking to understand the logic and math related to the calculation of heat loss of water in an insulated tube. My goal is to determine what inlet water temperature is required to avoid the water freezing before exiting the tube, but rather than just plugging in numbers, I want to understand the calculations.
The parameters:
• Tubing length: 10m
• Tubing size: 1/4" OD polyethylene (1/6" ID) "ice maker tubing"
• Inlet temperature: 5°C
• Flow rate: 3 l/min constant flow
• Ambient temperature: -25°C (indoors: no wind)
• Tubing thermal conductivity: ~0.4 W/m per °C
• Tubing insulation: 1/2" thick solid closed cell polyurethane rubber insulation sleeve (thermal conductivity: ~0.030 W/(m·K))
How would I calculate the thermal losses under these conditions to determine if it would freeze before exiting the tube, and also to determine if/how much heat needs to be added (i.e. raise the inlet temperature) to prevent freezing?
I'm not sure of the correct approach, but my guess is to calculate the thermal conduction of the tube + insulation per meter, then extrapolate to the full surface area of the tube to determine heat loss of the entire tube, then calculate if the water will reach 0°C at this heat loss rate. However, I'm confused on some points:
• How is flowing water accounted for vs. non-flowing water? Do I need to determine speed of flow to determine if the water will exit the tube before losing enough heat to reach 0°C? (i.e. how long is any particular water molecule present inside the tube)
• How is the thermal conductivity of the insulation combined with the thermal conductivity of the tube itself?
• I don't understand how to account for temperature gradient (i.e. water entering the tube being warmer than water exiting the tube). Must this be taken into account, or can the entire system be viewed as being a single temperature?
• Once I have a formula for heat loss, would I simply plug in a temperature slightly higher than 0°C as the desired output temperature, to calculate the inlet temperature required to avoid freezing?
Thanks for any pointers.
• The equations to for this are empirical, that is, determined via experiment. Speed of flow absolutely matters. This is what most of the ME fluids course is about. Jan 23 at 22:23
• Look up "forced convection in an insulated pipe" Jan 24 at 1:39
• Look up conduction in pipes. Jan 24 at 7:19
• @TigerGuy I make the Reynolds number of this pipe flow about $800$, i.e. in the laminar range, so it can be solved analytically. Jan 24 at 16:34
It sounds to me like a calculus problem if you want an accurate result but we might be able to simplify it.
From the information given you can calculate the water velocity in the pipe and from that you can calculate the time in the pipe.
• 3 L/minute = 3000 cm3/minute = 50 cm3/s.
• 1/6" diameter gives a cross-sectional area of 0.816 cm2.
$$v = \frac {flow\ rate}{cross\ sectional\ area} = \frac {50\ \mathrm{[cm^3/s]}} {0.816 \mathrm {[cm^2]}} = 61\ \mathrm{cm/s}$$ $$Time\ in\ tube = \frac L v = \frac {1000\ \mathrm {[cm]}}{61\ \mathrm {[s]}} = 16.4\ \mathrm s$$
Now you should be able to calculate the heat loss in that time. I'm an electrical engineer so it looks to me like a classic RC exponential decay circuit.
• The ΔT between the water and ambient will be equivalent to voltages.
• The thermal energy of the water will be equivalent to the capacitor charge.
• You'll need to calculate the total thermal resistance then and apply the exponential decay equation.
I might have a look at this another time if you can't figure it out and nobody else offers any help.
| 896
| 3,656
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.640625
| 4
|
CC-MAIN-2024-22
|
latest
|
en
| 0.922648
|
https://graphicallinearalgebra.net/tag/isomorphism/
| 1,680,050,225,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948900.50/warc/CC-MAIN-20230328232645-20230329022645-00356.warc.gz
| 324,604,949
| 36,931
|
# 19. Integer Matrices
In the last episode we introduced the fifth and final principal actor of graphical linear algebra, the antipode. This episode’s main task is showing that diagrams built up of the five generators constitute a diagrammatic language for integer matrices and their algebra. We will also discuss a cute example involving the complex numbers.
The cheat sheet for the diagrammatic system H that includes the antipode is repeated below for easy reference.
We have already showed that H allows us to extend the syntactic sugar for natural numbers to a sugar for all the integers. We have also verified that the integer sugar obeys the usual algebraic laws of integer arithmetic. In particular, we proved that
using the equations of H and diagrammatic reasoning: this is just -1 ⋅ -1 = 1, expressed with diagrams.
Let’s start with an example. The diagram below has two dangling wires on both sides.
As such, it ought to denote some 2×2 matrix, and to get the entry in the ith row and jth column we need to look at the paths from the jth dangling point on the left to the ith dangling point on the right.
Before the antipode entered the frame, it sufficed to count the number of paths; the current situation is a little bit more complicated because there are positive paths—those on which the antipode appears an even number of times—and negative paths, those with an odd number. To get the relevant integer entry, we need to take away the negative paths from the positive paths. So, in the very simple example above, we have exactly one positive path from the first point on the left to the second point on the right, and one negative path from the second point on the left to the first on the right. The corresponding matrix is therefore:
Actually, I didn’t choose this matrix at random: it allows us to consider the complex integers (sometimes called the Gaussian integers) and their algebra in a graphical way. We will come back to this after tackling the main topic for today.
We want to prove that H is isomorphic to the PROP MatZ of matrices with integer entries. The letter Z is often used to mean the integers, from the German word Zahl meaning number; this notation was apparently first used by Bourbaki.
MatZ is similar to the PROP Mat that we discussed in Episodes 12 and 13: the arrows from m to n are n×m matrices, and just like before composition is matrix multiplication. The monoidal product is again direct sum of matrices.
The proof H ≅ Mat(the symbol is notation for isomorphic to) is similar, and not much more difficult than the proof, outlined in Episodes 15 and 16, of ≅ MatLet’s go through the details.
First we define a homomorphism of PROPs from H to MatZ. Let’s call it φ, the Greek letter phi. Since both H and MatZ are PROPs, and H is a free PROP built from generators and equations, it is enough to say where φ sends all the generators, and then check that the equations of H hold in MatZ.
It turns out that φ works the same way as θ for all of the old generators. The new part is saying where the antipode goes, and not surprisingly, it is taken to the 1×1 matrix (-1). Like so:
For φ to be well-defined, we need to check that all the equations of H also hold in MatZ. Fortunately, most of that work was already done for θ, we only really need to check the new equations that involve the antipode. Let’s check the most interesting of these, (A1); we need to calculate whether the following is true:
This amounts to checking if
and it does, indeed, work. The other equations, (A2) through to (A5) are similarly easy computations and we will skip them; but feel free to check!
So we have a homomorphism φ: H → MatZ. To show that it is an isomorphism, we will show that it is full and faithful. Fullness—the fact that every matrix has a diagram that maps to it via φ—is the the easy part.
First, we need to check that the sugar that we defined in the last episode works with φ as expected, which is confirmed by the following simple calculation:
Any matrix with integers as entries can now be constructed following the procedure described in Episode 15. We will skip the details, as it is all pretty straightforward! The upshot of the construction is that we can extend the sugar for natural number matrices to a sugar for integer matrices: given an m×n integer matrix U we obtain a sugar
such that
This establishes that φ is full.
So what about faithfulness, the property that says that whenever two diagrams map to the same matrix then they must already be equal as diagrams?
The trick is to get our diagrams into the form where the
copying comes first, then the antipodes, then the adding (★)
One way of doing this is to use the theory of distributive laws. Eventually we will go through all of this properly, but for now I will just give you a high-level executive overview. The main insight is that we have three different distributive laws, the first involving the adding and the copying (B1)-(B4), the second the antipode and copying (A2)-(A3), and the third the antipode and adding (A4)-(A5).
The three distributive laws, are compatible with each other in a sense identified by Eugenia Cheng in her paper Iterated distributive laws. The fact that the distributive laws play together well in this way gives us the factorisation (★) that we want. We will discuss Cheng’s results in more detail in a later episode. Incidentally, she has recently written a book about category theory and recipes; I wonder if she knows about Crema di Mascarpone!
We could also try a rewriting argument, taking for granted that the rewriting system described in Episode 16 terminates. Adding the following rules
it seems that the expanded system ought to terminate also, although I have not yet got around to proving it. These termination proofs are always really messy for a rewriting amateur like me; I would love to hear from an expert about how to do these kinds of proofs in a nice way.
Once we know that every diagram can be put in the form (★), the proof of faithfulness is fairly straightforward. We start with those diagrams that have one dangling wire on each side. Every such diagram in the form (★) is either the sugar for 0 (a single discard followed by a single zero) or it can be rearranged into the form:
for some natural number k of wires with one antipode and some natural number l of wires with no antipode. This is because we can always get rid of redundant discards and zeros with (Counit) and (Unit), cancel out multiple antipodes in series using (†), then rearrange, and eat up any annoying permutations with the iterated copy and add sugars.
Once our diagram is in this form we can desugar and repeatedly use (A1), each time destroying one pair of antipode wire and no-antipode wire. Either we end up with no antipodes left, in which case the diagram is equal to a non-negative sugar, or we end up with some number of antipode wires. In the latter case, we can use (A2) to pull out the antipode to the left, obtaining the sugar for a negative integer. We have thus shown that faithfulness holds for the (1,1) case, since every such diagram is equal to some integer sugar.
The general case, where diagrams can have any number of wires on the left and right, comes down to transforming the diagram in matrix form, as explained in Episode 16. This step completes the proof that φ is faithful, and since we already know it is full, it is an isomorphism.
So far we have been identifying “numbers” with diagrams of a particular kind; those with one dangling wire on each end. In B this gave us the natural numbers, and in H it gives us the integers. But, as argued in Episode 17, there’s nothing particularly special about (1, 1) diagrams; well, maybe apart from the fact that both in B and H composition for (1,1) diagrams turns out to be commutative. Our obsession with the (1, 1) case is due to history—the traditional way of doing matrix algebra means the concept of ‘number” comes first, then the concept of “matrix”.
The complex numbers are a nice example where it makes sense to consider “numbers” as something different than (1,1) diagrams A complex number can be written as an expression r + si where rs are numbers and i is a formal entity that behaves like a number, but with the mysterious property i= -1. The numbers and s are sometimes called, respectively, the real component and imaginary components. What is important for us is that to describe a complex number, it suffices to keep track of two ordinary numbers. Our intuition is that wires carry numbers, so it makes sense to carry a complex number with two wires, the first for the real piece, the second for the imaginary piece.
Now if we multiply a complex number r + si by i, we get (r + si)i = ri + sii = -s + ri. So what was the real component becomes the imaginary component, and the negative of what was the imaginary component becomes the real component. We have a diagram for that, and we have already seen it in this episode:
It thus makes sense to call this diagram i:
Now if we multiply r + si by an integer u, we get (r+si)u=ru + sui. So both the components are multiplied by u. We also have a diagram for that:
where on the right hand side we used the sugar for integers from the last episode.
For the rest of this section, to stop the proliferation of the digit 2 that clutters the diagrams, we will just draw the 2 wire using a thicker line, like so:
Now we can do some calculations. First if we compose the diagram for i with itself we get:
We can also show that i commutes with integers:
Following the general pattern of this blog, we can ask what kinds of diagrams one can construct using the following gadgets.
Using our standard box of tricks for reasoning about diagrams, it is not difficult to show that the diagrams with one thick wire on each side will, in general, be of the form:
Composing two such entities gives us
which is of course what you’d get if you multiplied out two complex integers (those complex numbers u+vi where u and v are integers). In general, the diagrams that can be constructed from bricks (‡) are matrices with complex integer entries.
So what exactly is going on here? Let’s take a look under the hood.
The result is in matrix form, and corresponds to the 2×2 matrix:
and this is known as one way of representing complex numbers using matrices.
There is one more interesting thing to say here. Let’s take a look at the bizarro of i.
So the bizarro of i is -i. It follows that the bizarro of a general diagram constructed in the system (‡) corresponds to the operation known as conjugate transpose in complex matrix algebra.
If you know about quaternions, they can be considered in a similar way. Of course, we are constrained to integer coefficients for now. Not for long ☺.
I will give a 3 hour tutorial about graphical linear algebra at QPL ’15 in two chunks on Monday and Tuesday of next week. I’m desperately trying to get the slides done on time. Running this blog has been helpful in that it forced me to develop material, but unfortunately what we have covered so far will only be enough for around the first 30 mins; I should have started this blog back in January!
Continue reading with Episode 20 – Causality, Feedback and Relations.
# 14. Homomorphisms of PROPs
In the last two episodes we have established that our diagrams, constructed from the four generators, and subject to the equations below, are the arrows of a PROP B.
We have also talked about another PROP, Mat, that has as its arrows matrices of natural numbers.
In maths, it is often the case that to understand something properly you need to think about how it relates to other things. Our current task is an example of this, since we need to establish exactly what we mean when we say that diagrams and matrices “are the same thing”. This brings us to the notion of homomorphism —a generic jargon word that means “structure-preserving translation”—of PROPs.
So suppose that X and Y are PROPs. It’s useful to colour them so that we can keep track of what’s going on in which world, the red world of X and the hot pink world of Y. A homomorphism F from X to Y (written F : X Y) is a function (a translation) that
for any arrow : m → n in X produces an arrow F: m → n in Y.
Notice that the domain and the codomain must be preserved by the translation: both A in X and FA in Y have domain m and codomain n.
But PROPs come with a lot of structure, and homomorphisms have to preserve all of it. For example, for any other arrow A’ in X, it must be the case that
F(A A’) = FA FA’
The above equation says that if I translate the monoidal product of A and A’ taken in X then I get the same thing as if I had translated A and A’ separately and took the monoidal product in Y. Remember that monoidal product is defined differently in different PROPs: for example in B it is the stacking diagrams on top of each other, while in Mat it is forming a certain matrix.
Similarly, if we can compose A with A’ (which we can do exactly when the domain of A’ is n) then:
F(A ; A’) = F ; FA’ ②
Again this says that translating A composed with A’ in X should be the same thing as translating the individual components separately and doing the composition in Y. Recall that in B composition is simply connecting dangling wires together, while in Mat it is matrix multiplication.
Moreover, as we have seen, every PROP has special identity arrows, one for each natural number m. These form a part of the structure of PROPs and so must also be preserved:
Fidm = idm ③
Finally, for any natural numbers m, n, there are the special “wire-crossing” arrows σm,n that also must be preserved:
Fσm,n = σm,n ④
That takes care of all the structure. So if we can find a translation F that satisfies and , we have got ourselves a homomorphism of PROPs.
Back in Episode 11 we started to define a translation called θ. There we listed the matrix that is assigned to each generator:
But to have a homomorphism from B to Mat, we need to give a rule θ that translates every diagram with m dangling wires on the left and n dangling wires on the right to an n×m matrix. It turns out that because of the way in which B is constructed, the above four rules tell us everything we need in order to define a homomorphism from B to Mat. All the other stuff is predetermined.
This is because B is special: it has diagrams as arrows, and every diagram is constructed from the four generators, identity and twist. That means that there is only one way we can extend the translation above so that it defines a homomorphism. We do not even need to say how to translate the identity and twist diagrams, since they are part of the PROP structure: they are respectively id1 and σ1,1 of the PROP B. And together with tell us that they must be translated to id1 and σ1,1 in the PROP Mat:
Next, since we already know how to translate the most basic diagrams, rules and give us the recursive rules for translating more complex diagrams. We already worked through an example.
Because of the middle-four interchange and the identity laws it does not matter how we subdivide the diagram: since Mat is also a PROP, it also satisfies these laws. Moreover, the procedure is well-defined, since, as we have seen in the last two episodes, diagrammatic reasoning amounts to using equations that hold in any PROP, and the additional equations of B that can be shown to hold in Mat: we did the work of checking this back in Episode 11.
That alleviates all of our fears from when we started discussing the translation. All that hard work of going through PROP lore is finally paying off! If you are not that impressed then don’t worry, we will have the opportunity to see many other examples of PROPs in action further on in the story.
Stepping back a bit from what we’ve been doing above, there is one thing worth mentioning. Back when we were talking about how natural numbers can be seen as special kinds of diagrams, I mentioned that they are free: two diagrams are considered to be equal exactly when we can prove it using diagrammatic reasoning and our equations.
In fact, the PROP of diagrams is the free PROP on the four generators and ten equations. This is the “special” nature of B that I alluded to before, and the thing that makes PROP homomorphism from B pretty easy to define: we just need to show where the generators go and make sure that the equations of B are also true in the codomain PROP.
How can we tell, in general, that two different languages can express exactly the same concepts?
One way is to construct a perfect translation. To understand what this could mean, we could start by thinking what non-perfect translations look like. For example, let’s pretend that I claim to be qualified English to French translator—which I’m definitely not, by the way—and you give me two English words to translate, say serendipity and dog. Suppose that both of the times I say chien. Then my translation cannot be very good, even if you don’t know what the word chien means.
The reason is that you could ask me to translate chien back to English. This would force me to choose one of serendipity or dog, or maybe something else entirely. If I say dog then you know that something went wrong: you asked me to translate serendipity to French, I said chien, then you asked me to translate back and I said dog.
serendipity → chien → dog
Since you know that serendipity and dog are two different concepts, something clearly got lost in the translation. Even If I said serendipity, you would still be able to catch me out, since then the translation chain would be:
dog → chien → serendipity
The moral of this story is that we would expect that a reasonable translator would not translate two different concepts in English to the same word in French.
The mathematical jargon adjective for such reasonable translations is injective. And because PROPs come from category theory, they inherit their own special jargon: a homomorphism F : X Y is said to be faithful when,
given two different arrows A A’ : m → n in X, we have FA FA’: m → n in Y.
Another way a translation, say from Klingon to English, could be less than satisfactory is if there are some words in English for which a word in Klingon does not exist. This is likely because Klingon only has about 3000 words: so some English words like “serendipity” do not have a Klingon equivalent. But don’t quote me on that, I’m not an expert in Klingon.
The common mathematical jargon for a translation that hits every word in the target language is surjective, and in the world of PROPs the word is full. So a PROP homomorphism F: X Y is said to be full when
for all arrows B: m → n in Y, there exists an arrow A: m → n in X such that FA = B.
It turns out that a homomorphism F: X Y that is both full and faithful is perfect in the the following sense: there exists a translation G: Y X that “reverses” F. G is called the inverse of F and satisfies the following two properties:
1. for all arrows A: m → n in X, GFA = A
2. for all arrows B: m → n in Y, FGB = B
So if I start with some A in X, translate it to Y and translate back again I end up where I started. Same if I start with some B in Y. There is a special word for homomorphisms that have inverses: they are called isomorphisms. A translation that has an inverse is about as perfect as one could expect.
The PROP homomorphism θ: Mat is full and faithful and thus an isomorphism of PROPs. We will discuss why this is the case in the next episode.
The upshot of θ being an isomorphism is that the diagrams of B and matrices of natural numbers are really two languages that talk about the same thing. In particular, we should be able to translate concepts from one language to the other.
Here’s one example: we saw that the bizarro operation on diagrams, where we reflect a diagram and interchange white and black, is quite useful: it has already saved us quite a bit of work with proofs, since a proof of any claim can always be translated to a bizarro proof of the bizarro claim. So what does it mean to consider the bizarro version of a matrix: that is, if I start with a diagram D and its bizarro version Dbizarro, what is the relationship between matrices θD and θDbizarro?
Well, it turns out that the equivalent concept for matrices is called transpose. If I give you an m×n matrix A (n columns, m rows) then its transpose AT is an n×m matrix (m columns, n rows) that has, as its entry at the ith row and jth column, the entry at the jth row and the ith column of A. Intuitively speaking, the rows of A become the columns of AT. Here’s an example, if I let A be the matrix
then its transpose is the matrix
What about a concept that maybe you’ve come across in the language of matrices? For example, linear algebra courses usually go on about special kinds of matrices called row vectors and column vectors. A row vectors is simply a matrix with exactly one row, and a column vector is a matrix with exactly one column.
So the concept of a row vector, translated to the world of diagrams, is a diagram with exactly one dangling wire on the right. Here’s an example:
Similarly, a column vector translates to a diagram with exactly one dangling wire on the left. Like so:
Some of you, especially those who are already familiar with matrices, are probably asking yourselves what is the point of having two languages to describe the same thing. It all seems to be a bit redundant, since you already know about the concept of a matrix of natural numbers. Please hang in there for now: I hope to convince you that looking at the world through the prism of diagrams gives you a different, sometimes truly surprising perspective.
Continue reading with Episode 15 – Matrices, diagrammatically
| 5,135
| 21,824
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875
| 4
|
CC-MAIN-2023-14
|
latest
|
en
| 0.884337
|
https://ask.sagemath.org/question/53429/solving-a-diophantine-system/?sort=latest
| 1,721,161,267,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514789.44/warc/CC-MAIN-20240716183855-20240716213855-00513.warc.gz
| 90,773,615
| 13,911
|
# Solving a Diophantine system
I am trying to find solutions to the following Diophantine system:
a+b+c=x^2
a^2+b^2+c^2=y^2
a^3+b^3+c^3=z^2
where a,b,c are less than 5000 and where x,y,z are perfect squares
edit retag close merge delete
Should $x$, $y$, $z$ really be perfect squares?
So $a + b + c$, $a^2 + b^2 + c^2$, $a^3 + b^3 + c^3$ are perfect fourth powers?
Or did you mean $a + b + c$, $a^2 + b^2 + c^2$, $a^3 + b^3 + c^3$ are perfect squares?
If so, write one of the following:
• $a + b + c = x$, $a^2 + b^2 + c^2 = y$, $a^3 + b^3 + c^3 = z$ and $x$, $y$, $z$ are perfect squares
• $a + b + c = x^2$, $a^2 + b^2 + c^2 = y^2$, $a^3 + b^3 + c^3 = z^2$ and nothing more on $x$, $y$, $z$
( 2020-09-13 03:43:41 +0200 )edit
Is this homework? Is this from Project Euler?
What have you tried? What specific problem(s) are you facing?
( 2020-09-13 03:45:22 +0200 )edit
It is not for homework, I am doing research and I am just trying to find small numerical solutions without using elliptic curves.
I do mean that $a+b+c$, $a^2+b^2+c^2$, and $a^3+b^3+c^3$ are simply perfect squares, not perfect fourths.
And i have been trying to define each as a separate equation but I cant solve for the variables within a range as integers, something like:
for a in range(5000) for b in range(5000) for c in range(5000)
eq1 = a+b+c==x^2 eq2 = a^2+b^2+c^2==y^2 eq3 = a^3+b^3+c^3==z^2
solve([eq1,eq2,eq3],a,b,c,x,y,z)
( 2020-09-13 09:01:45 +0200 )edit
Sort by ยป oldest newest most voted
Elementary version
sage: for a in srange(1, 5000):
....: for b in srange(1, a+1):
....: for c in srange(1, b+1):
....: if (a+b+c).is_square() and (a^2+b^2+c^2).is_square() and (a^3+b^3+c^3).is_square():
....: print(a,b,c)
A little bit more elaborate
sage: for abc in srange(1, 10000):
....: abc = abc*abc
....: for a in srange((abc + 2)// 3, abc):
....: for b in srange((abc - a + 1)//2, min(a, abc - a)):
....: c = abc - a - b
....: if (a^2+b^2+c^2).is_square() and (a^3+b^3+c^3).is_square():
....: print(a,b,c)
From which you get the list: (129, 124, 108), (516, 496, 432), (1161, 1116, 972), (2873, 2134, 34), (2064, 1984, 1728), (3225, 3100, 2700), ...
more
Thank you so much! It is exactly what I needed!!
( 2020-09-13 10:35:18 +0200 )edit
| 902
| 2,298
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.953125
| 4
|
CC-MAIN-2024-30
|
latest
|
en
| 0.734379
|
http://lolengine.net/wiki/research/trig?action=diff&version=6
| 1,603,276,701,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107876307.21/warc/CC-MAIN-20201021093214-20201021123214-00127.warc.gz
| 62,703,790
| 4,513
|
Changes between Version 5 and Version 6 of research/trigTweet
Ignore:
Timestamp:
Oct 13, 2011, 4:42:46 PM (9 years ago)
Comment:
use \max instead of \forall
Unmodified
Added
Removed
Modified
• research/trig
v5 {{{ #!latex $\big\vert\sin(x) - P(x)\big\vert \le E \qquad \forall x \in \bigg[-\frac{\pi}{2}, \frac{\pi}{2}\bigg]$ $\max_{x \in [-\pi/2, \pi/2]}{\big\vert\sin(x) - P(x)\big\vert} = E$ }}} {{{ #!latex $\big\vert\sin(x) - xQ(x^2)\big\vert \le E \qquad \forall x \in \bigg[-\frac{\pi}{2}, \frac{\pi}{2}\bigg]$ $\max_{x \in [-\pi/2, \pi/2]}{\big\vert\sin(x) - xQ(x^2)\big\vert} = E$ }}} {{{ #!latex $\big\lvert\sin(\sqrt{y}) - \sqrt{y}Q(y)\big\rvert \le E \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\big\lvert\sin(\sqrt{y}) - \sqrt{y}Q(y)\big\rvert} = E$ }}} {{{ #!latex $\bigg\lvert\frac{\sin(\sqrt{y})}{\sqrt{y}} - Q(y)\bigg\rvert \le \frac{E}{|\sqrt{y}|} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})}{\sqrt{y}} - Q(y)\bigg\rvert}{\dfrac{1}{|\sqrt{y}|}}} = E$ }}} If we want to force the asymptotic behaviour in x=0, we substitute Q(y) with 1+yR(y): If we want to force the asymptotic behaviour at x=0, we substitute Q(y) with 1+yR(y): {{{ #!latex $\bigg\lvert\frac{\sin(\sqrt{y})}{\sqrt{y}} - 1 - yR(y)\bigg\rvert \le \frac{E}{|\sqrt{y}|} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})}{\sqrt{y}} - 1 - yR(y)\bigg\rvert}{\dfrac{1}{|\sqrt{y}|}}} = E$ }}} {{{ #!latex $\bigg\lvert\frac{\sin(\sqrt{y})-\sqrt{y}}{y\sqrt{y}} - R(y)\bigg\rvert \le \frac{E}{|y\sqrt{y}|} \qquad \forall y \in \bigg[0, \frac{\pi^2}{4}\bigg]$ $\max_{x \in [0, \pi^2/4]}{\dfrac{\bigg\lvert\dfrac{\sin(\sqrt{y})-\sqrt{y}}{y\sqrt{y}} - R(y)\bigg\rvert}{\dfrac{1}{|y\sqrt{y}|}}} = E$ }}}
| 837
| 1,857
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.59375
| 4
|
CC-MAIN-2020-45
|
latest
|
en
| 0.352796
|
http://www.meracalculator.com/unitconverter/electric-field-strength-intensity.php
| 1,553,244,195,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-13/segments/1552912202640.37/warc/CC-MAIN-20190322074800-20190322100800-00342.warc.gz
| 331,437,497
| 3,857
|
# Electric Field Strength Converter
Electric field refers to a region around a charged particle or object within which a force would be exerted on other charged particles or objects. An electric field that changes with time, such as due to the motion of charged particles in the field, influences the local magnetic field.
The electric field intensity is defined as the force per unit positive charge that would be experienced by a stationary point charge, or "test charge", at a given location in the field.
Electric Field Strength, Intensity Converter
Enter Value: --Select-- Abvolt/centimeter Kilovolt/centimeter Microvolt/meter Millivolt/meter Statvolt/centimeter Statvolt/inch Volt/centimeter Volt/inch Volt/meter Abvolt/centimeter: abV/cm Kilovolt/centimeter: kV/cm Microvolt/meter: µ/m Millivolt/meter: mV/m Statvolt/centimeter: stV/cm Statvolt/inch: stV/in Volt/centimeter: V/cm Volt/inch: V/in Volt/meter: V/m
Formula
V/m = (W/m2 x 377)1/2
The electric field is a vector field with& SI units of newtons per coulomb (N C−1) or, equivalently, volts per metre (V m−1). The SI base units of the electric field are kg·m·s−3·A−1.
Example:
Find Electric Field Strength, intensity conversion.
Enter value: 5 Abvolt/centimeter
Solution:
Abvolt/centimeter = 5
Kilovolt/centimeter = 4.9999999999999995e-11
Microvolt/meter = 5
Millivolt/meter = 0.005
Statvolt/centimeter = 1.667823024e-10
Statvolt/inch = 4.2362704808650005e-10
Volt/centimeter = 5e-8
Volt/inch = 1.2699999999999999e-7
Volt/meter = 0.0000049999999999999996
| 447
| 1,525
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.140625
| 3
|
CC-MAIN-2019-13
|
latest
|
en
| 0.795252
|
https://teterfouband.firebaseapp.com/117.html
| 1,670,385,218,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711126.30/warc/CC-MAIN-20221207021130-20221207051130-00257.warc.gz
| 582,203,709
| 6,155
|
# Ring theory mathematics pdf
Most parts of algebra have undergone great changes and advances in recent years, perhaps none more so than ring theory. Mathematics rings, integral domains and fields geeksforgeeks. In mathematics, a ring is one of the fundamental algebraic structures used in abstract algebra. The natural numbers, n are what number theory is all about. Abstract algebra sydney mathematics and statistics. Groups play an important role nearly in every part of mathematics and can be.
Whether a ring is commutative or not that is, whether the order in which two elements are multiplied changes the result or not has profound implications on its behavior as an abstract object. Lecture notes modern algebra mathematics mit opencourseware. Pdf on oct 17, 2019, akeel ramadan mehdi and others published ring theory find, read and cite all the research you need on researchgate. On the other hand the presentation includes most recent results and includes new ones. Ring theory and its applications ring theory session in honor of t. This set of notes is based on the course introduction to string theory which was taught by prof. Thanks for contributing an answer to mathematics stack exchange. In algebra, ring theory is the study of rings algebraic structures in which addition and multiplication are defined and have similar properties to those operations defined for the integers. For the literary technique, see chiastic structure. So, a group holds four properties simultaneously i closure, ii associative, iii identity element, iv inverse element. Although people have been studying specific examples of rings for thousands of years, the emergence of ring theory as a branch of mathematics in its own right is a very recent development.
Dabeer mughal federal directorate of education, islamabad, pakistan. This means that in mathematics, one writes down axioms and proves theorems from the axioms. The foundations of mathematics involves the axiomatic method. Feb 18, 2018 51 videos play all ring theory arvind singh yadav,sr institute for mathematics integral domains abstract algebra duration. Matsumura covers the basic material, including dimension theory, depth, cohenmacaulay rings, gorenstein rings, krull rings and valuation rings. The study of rings has its roots in algebraic number theory, via rings that are generalizations and extensions of. Group theory notes michigan technological university. Introduction to ring theory sachi hashimoto mathcamp summer 2015 1 day 1 1. In many ways it will look like our familiar notions of addition and multiplication, but sometimes it wont. Rings, properties of rings, integral domains and fields, subrings, idempotent and nilpotent elements, characteristic of a ring, ideals in a ring, simple ring, homomorphisms, principal ideal domains, euclidean domains, polynomial rings, unique factorization domain, extension fields.
Foundations of module and ring theory download book. I would suggest you go through the following steps. And make certain that, when you use mathematical symbols. Consider a set s nite or in nite, and let r be the set of all subsets of s. Interpreting matdumura as a function that takes the value f mod p i. Introduction to groups, rings and fields ht and tt 2011 h. Starting from a basic understanding of linear algebra the theory is presented with complete proofs. Recommended problem, partly to present further examples or to extend theory. Ring theorists study properties common to both familiar mathematical structures such as integers and polynomials, and to the many less wellknown mathematical structures that also satisfy the axioms of ring theory.
Ring theory studies the structure of rings, their representations. If gis a group of even order, prove that it has an element a6esatisfying a2 e. Field a nontrivial ring r wit unity is a field if it is commutative and each nonzero element of r is a unit. Whereas ring theory and category theory initially followed di. But avoid asking for help, clarification, or responding to other answers. In this paper, we propose a new index of similarity among images using. However, a student of algebra and many other branches of mathematics which use algebra in a nontrivial way, e. Ring theory is generally perceived as a subject in pure mathematics. An introduction to string theory kevin wray abstract. The ubiquity of rings makes them a central organizing principle of contemporary mathematics. Lam on his 70th birthday at the 31st ohio statedenison mathematics conference, may 2527, 2012, the ohio state university, columbus, oh dinh van huynh, s. An early contributor to the theory of noncommutative rings was the scottish mathematician wedderburn, who in 1905, proved wedderburns theorem, namely.
Broadly speaking, a ring is a set of objects which we can do two things with. Much of the activity that led to the modern formulation of ring theory took place in the first half of the 20th century. This is an abridged edition of the authors previous twovolume work, ring theory, which concentrates on essential material for a general ring theory course while ommitting much of the material intended for ring theory specialists. These notes are aimed at students in the course ring theory mat 3143 at the university of ottawa. Now for any a2gwe have ea ayaa ayaa ae aas eis a right identity. Commutative ring theory cambridge studies in advanced.
Dabeer mughal a handwritten notes of ring algebra by prof. What is ring theory and group theory in mathematics. Questions tagged ring theory ask question this tag is for questions about rings, which are a type of algebraic structure studied in abstract algebra and algebraic number theory. Open problems in commutative ring theory pauljean cahen, marco fontanay, sophie frisch zand sarah glaz x december 23, 20 abstract this article consists of a collection of open problems in commutative algebra.
However, ring theory has not been very related with image segmentation. This introductory section revisits ideas met in the early part of analysis i and in linear algebra i, to set the scene and provide. A ring in which x2 xfor all elements is called a boolean ring. If gis a nonempty set, a binary operation on g is a function. We have also drawn on some ideas from the books string theory and. The inclusion of ring theory to the spatial analysis of digital images, it is achieved considering the image like a matrix in which the elements belong to finite cyclic ring. It is known in basic ring theory that any ring r with identity can be embedded in its own endomorphism ring endr this is the analogue of the cayley theorem in group theory. The collection covers a wide range of topics from both noetherian and nonnoetherian ring theory and exhibits a variety of re.
Commutative ring theory is important as a foundation for algebraic and complex analytical geometry and this text covers the basic material with a solid knowledge of modern algebra as the only prerequisite. The more recent version is called commutative ring theory and is still in print. Ring theory math berkeley university of california, berkeley. The inverse element denoted by i of a set s is an element such that a. In this way the book will prove stimulating to those. A ring is a set equipped with two operations usually referred to as addition and multiplication that satisfy certain properties. Ring theory has been wellused in cryptography and many others computer vision tasks 18. A slender volume and not very comprehensive but you can cut to the heart of the matter rather quickly. Ring theory math 1, summer 2014 james mcivor university of california, berkeley august 3, 2014 abstract. In this paper, we propose a new index of similarity among images using rings and. Assume to the contrary that iis not a subset of jand that jis not a subset of i. Ring theorists study properties common to both familiar mathematical structures such as integers and polynomials, and to the many less wellknown. Without going into too much detail, just as an integral domain is the most faithful abstraction of the ring of integers, a noetherian ring is the best way to think about polynomial rings and. The branch of mathematics that studies rings is known as ring theory.
This means that it is a subject of intrinsic beauty. Exercises and solutions in groups rings and fields 5 that yaayat ethen yaae e hence yaa e. Ring theory math 1, summer 2014 james mcivor university of california, berkeley august 3, 2014 abstract these are some informal notes on rings and elds, used to teach math 1 at uc berkeley, summer 2014. In addition to being an interesting and profound subject in its own right, commutative ring theory is important as a foundation for algebraic geometry and complex analytical geometry. Kostas skenderis in the spring of 2009 at the university of amsterdam. In math, we often look at homomorphisms to tell us more about the structure of a ring, rather than looking at the individual ring itself. Lam on his 70th birthday 31st ohio statedenison mathematics conference may 2527, 2012 the ohio state university, columbus, oh dinh van huynh s. After a chapter on the definition of rings and modules there are brief accounts of artinian. In contrast to commutative ring theory, which as we have seen grew from number theory, noncommutative ring theory developed from an idea which, at the time of its discovery, was heralded as a great advance in applied mathematics. It consists of a set equipped with two binary operations that generalize the arithmetic operations of addition and multiplication. The order of a group g is the number of elements in g and. If every x2rsatis es x2 x, prove that rmust be commutative.
As a result, commutative ring theory, commonly known as commutative algebra, is a key topic in ring theory. In our presentation many of the results obtained this way are achieved by. X x iaor b x 1x 2 x m 1x m thus the expression is equally valid for n m. Galois introduced into the theory the exceedingly important idea of a normal subgroup, and the corresponding division of groups into simple. This is an example of a quotient ring, which is the ring version of a quotient group, and which is a very very important and useful concept. Questions tagged ringtheory ask question this tag is for questions about rings, which are a type of algebraic structure studied in abstract algebra and algebraic number theory. Newest ringtheory questions mathematics stack exchange. Unlike for general rings, for a principal ideal domain, the properties of individual elements are strongly tied to the properties of the ring as a whole. Our usual addition and multiplication over the set of integers is. Ring theory is one of the branches of the abstract algebra that has been broadly used in images. Cambridge studies in advanced mathematics book 8 paperback. Each section is followed by a series of problems, partly to check understanding marked with the letter \r. Developments in pure mathematics the theory of rings structures in which it is possible to add, subtract, and multiply but not necessarily divide was much harder to formalize. There must also be a zero which functions as an identity element for addition, negatives of all elements so that adding a number and its negative.
Through this generalization, theorems from arithmetic are extended to nonnumerical objects such as polynomials, series, matrices and functions. Roman editors american mathematical society providence, rhode island. The concept of a noetherian ring is a particularly juicy one, and it was made famous by the founding mother of commutative ring theory, emmy noether. Browse other questions tagged abstractalgebra ringtheory or ask your own question.
1252 890 1269 629 1300 288 864 1069 522 87 1080 126 1415 954 369 1360 675 82 916 1427 1457 587 1472 500 477 624 740 478
| 2,523
| 11,847
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2022-49
|
latest
|
en
| 0.924726
|
juliettesierra.de
| 1,685,632,636,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224647895.20/warc/CC-MAIN-20230601143134-20230601173134-00426.warc.gz
| 27,511,636
| 4,758
|
Programming and
Design by
Jens Schöbel
# Self-Explanatory Code
Here I show you how to avoid comments in code. Some might think comments are needed to understand the according source. Unfortunately some problems will arise if you change your code but you forget to change the comment. Comments are good if they mean what they say. Hence the question come up, how to write code that avoids this issue?
## The Example
Lets start with the following piece of code. Thats a snipped of code I had to deal with as I was a scientific member of the University of Heidelberg. Everything is written in plain C without classes. Why nobody was using object orientation I don't know. I assume its just the lack of experience since most programmers are beginners at the university. I started changing this section as I run into trouble with this:
void DetermineAreas (void) { ... // Determines the area of a rectangular Area1 = DetermineArea(0, side1); // Determines the area of a rectangular Area2 = DetermineArea(0, side2, side3); ... }
You might think whats wrong with this. I tell you. Look at the first comment. Defined was DetermineArea(...) as:
float DetermineArea( int t, float l1, float l2 = -1.f); float DetermineArea( int t, float l1, float l2) { float Area = 0; // if no second argument given its a square if (l2 < 0) { l2 = l1; t = 1; } switch (t) { case 0: // Area of a rectangle Area = l1 * l2; break; case 1: // Area of a square Area = l1 * l1; break; default: // error printf ( "Error" ); break; } return Area ; }
DetermineArea(...) is very convoluted. Before you start wondering; this code isn't too complex. Its still understandable and thats why I've choosen this piece. Now lets have a closer look.
### Problems:
1. The mentioned comment will tell you that you are calculating the area of a rectangular ... or a square ... Who knows? You allways can say a square is a special rectangular but thats not what I mean.
2. DetermineArea(...) uses an optional Parameter. I don't like that since you allways have to look into the definition of a mehtod to find out what its doing.
3. I don't like the type parameter t. Since you need to look into the source to find out what's going on here.
Lets do some clean ups. But before this a warning. Please don't do clean ups without test. Sometimes you need to deal with the fact that there are no auto tests. Thats bad. And now guess how much code was covered with tests.
### Refactor steps:
1. Create an enum EAreaType and use this instead of t.
2. This way the first comment becomes obsolete. EAreaType will tell you the type.
3. If you have a closer look at the optional parameter you see that if l2 is missing its the same as l1. According to this you make l2 mandatory and change all the function calls - thats easy the compiler will tell you if something went wrong.
4. l1 and l2 aren't good names so lets change this to something more meaningful.
enum EAreaType { SQUARE = 0, RECTANGLE }; void DetermineAreas(void) { ... Area1 = DetermineArea(RECTANGLE, side1, side1); Area2 = DetermineArea(RECTANGLE, side2, side3); ... } DetermineArea( EAreaType type, float sideLength1, float sideLength2); float DetermineArea( EAreaType type, float sideLength1, float sideLength2) { float Area = 0; switch (type) { case SQUARE: // Area of a rectangle Area = l1 * l2; break; case RECTANGLE: // Area of a square Area = l1 * l1; break; default: // error printf ( "Error" ); break; } return Area ; }
Thats better but there is more you can do.
### Refactor steps:
1. In the first line of DetermineAreas(...) is a square calculated, hence change the type here.
2. A switch with code is a code smell. Therefore extract two methods for it.
3. To make code more readable you define an ERROR. Defining it will blow up the code but will make it much easier to maintain.
#define ERROR -1 void DetermineAreas (void) { ... Area1 = DetermineArea(SQUARE, side1, side1); Area2 = DetermineArea(RECTANGLE, side2, side3); ... } float DetermineArea ( EAreaType type, float sideLength1, float sideLength2) { switch (type) { case SQUARE : return GetAreaOfSqure(sideLength1); case RECTANGLE: return GetAreaOfRectangle(sideLength1, sideLength2); default: DrawError(); } return ERROR; } float GetAreaOfSqure (float sideLength1); { return sideLength1 * sideLength1; } float GetAreaOfRectangle( float sideLength1, float sideLength2) { return sideLength1 * sideLength2; }
Thats much better but still there is more you can do.
### Refactor steps:
1. If you have a closer look at the switch statement its easy to get rid of it. Instead of calling GetAreaOfSqure(...) through DetermineArea(...) you can call it derectly. The same holds for GetAreaOfRectangle(...).
2. By getting rid of the switch also EAreaType and ERROR become obsolete so delete it. This means they were just placeholders on the way to your final code. Now you might understand why I don't care if I blow up my code. Mostly you will delete lot of code later.
Thats it. Finaly you got a much cleaner code. With this little steps of refactoring you reduced your ammount of code and made it more readable by completely avoiding to write a single line of comment.
The code becomes self-explanatory now.
### After Refactoring:
void DetermineAreas (void) { ... Area1 = GetAreaOfSqure(side1); Area2 = GetAreaOfRectangle(side2, side3); ... } float GetAreaOfSqure(float sideLength1); { return sideLength1 * sideLength1; } float GetAreaOfRectangle( float sideLength1, float sideLength2) { return sideLength1 * sideLength2; }
| 1,462
| 5,753
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.3125
| 3
|
CC-MAIN-2023-23
|
latest
|
en
| 0.883371
|
https://www.topperlearning.com/doubts-solutions/icse-class-9-/volcanoes/
| 1,560,807,244,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-26/segments/1560627998580.10/warc/CC-MAIN-20190617203228-20190617225228-00384.warc.gz
| 933,390,950
| 59,071
|
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days
8104911739
or
Thanks, You will receive a call shortly.
Customer Support
You are very important to us
For any content/service related issues please contact on this toll free number
022-62211530
Mon to Sat - 11 AM to 8 PM
Ask ICSE Class 9 question free ×
Queries asked on Sunday and after 7 pm from Monday to Saturday will be answered after 12 pm the next working day.
# ICSE Class 9 Free Doubts and Solutions
## Naya Rasta ka book review kaise karenge
Asked by gennymariya83 16th June 2019, 1:29 PM
×
## What are the achievements India gained from Five year planning in different sectors?
Asked by amitrawat792 16th June 2019, 1:13 PM
×
## Summary of poem bhikshook from sahitya sagar.
Asked by ayush421301 16th June 2019, 10:16 AM
×
## evaluate
Asked by nandanam2020 15th June 2019, 7:36 PM
×
## 4x +5y =13 26x+3y =4 Please solve using cross multiplication method... Fast
Asked by krcraju1973 15th June 2019, 11:09 AM
×
## On the basis of which plan was the constituent assembly constituted
Asked by Seetesh_sawant 13th June 2019, 9:22 PM
×
## In the expansion of (2x^2-8) (x-4)^2 ;find the value of : i -cofficient of x^3, ii- cofficient of x^3 and constant term.
Asked by screen1974 12th June 2019, 8:46 PM
×
## Why we multiply a fraction say a/b with 100% to get the percentage.like a/b*100% ...explain with the help of example
Asked by Anshu 12th June 2019, 1:21 PM
×
## Arrange 5/8, 3/16, -1/4, and 17/32 in the descending order of their magnitudes. Also, find the sum of the lowest and the largest of these rational numbers express result obtained as a decimal fraction correct to two decimal places.
Asked by Sarojarongali 12th June 2019, 9:20 AM
## Answer this question
×
### ICSE - IX - Mathematics
Asked by ayushmani1871 10th June 2019, 11:00 AM
| 592
| 1,860
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.765625
| 3
|
CC-MAIN-2019-26
|
latest
|
en
| 0.933069
|
http://www.enotes.com/homework-help/you-shopping-fruit-market-you-want-spend-no-more-51559
| 1,462,154,294,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-18/segments/1461860121418.67/warc/CC-MAIN-20160428161521-00175-ip-10-239-7-51.ec2.internal.warc.gz
| 488,295,648
| 12,441
|
# You are shopping at the fruit market.You want to spend no more than \$12.00 on apples and bananas.Bananas cost \$2 per bunch,and apples cost \$4 per bagquestion:A.please write the inequality.B.find 3...
You are shopping at the fruit market.You want to spend no more than \$12.00 on apples and bananas.Bananas cost \$2 per bunch,and apples cost \$4 per bag
question:
B.find 3 possible combinations of bananas and apples that you can buy.write each combination as an ordered pair.let x be the number of banana bunches,and let y be the number of apple bags.
Posted on
The inequality is:
2x + 4y </= 12 [that should be < on top of _ , for less than or equal to, but I can't type that here]
so, x is the number of bunches of bananas (at \$2 each) and y is the number of bags of apples (at \$4 each). Since you can't spend more than \$12, the purchase must be less than or equal to 12.
Three possible combinations would be:
x = 1 and y = 2 (2*1 + 4*2 = 2 + 8 = 10)
x = 2 and y = 2 (2*2 + 4*2 = 4 + 8 = 12)
x = 3 and y = 1 (2*3 + 4*1 = 6 + 4 = 10)
Posted on
A. 2x + 4y < 12 or y < -1/2x + 3
remember x = bananas and y = apples
**Note all the < are < and = to.
Posted on
A. So this person wants to spend no more than \$12.00. The mouth of the inequality symbol will automatically point in the direction of \$12.00 with a equal to or less symbol.
\$12.00 >/
Now we know the cost of the bananas and apples so the equation to figure that out will go on the right so...
\$12.00 >/ \$2x+ \$4Ay
B Three possible combinations you could buy would be...
• 2 apples and 2 oranges
• 4 oranges and 1 apple
• 1 apple and 3 oranges
Posted on
2x+4y `<=` 12 dollars
Possible combinations:
1. 4 bananas and 1 apple
2. 1 apple and 3 bananas
3. 2 apples and 2 bananas
Posted on
The inequality is:
2x+4y less than or equal to 12 dollars
Possible scenarios:
1. Buy four bananas and 1 apple = 12 dollars
2. Buy two bananas and two apples= 12 dollars
3. Buy 1 banana and two apples= 10 dollars
| 613
| 2,005
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375
| 4
|
CC-MAIN-2016-18
|
longest
|
en
| 0.864401
|
https://ey.westside66.org/category/teachers/page/3/
| 1,675,055,088,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499801.40/warc/CC-MAIN-20230130034805-20230130064805-00518.warc.gz
| 267,299,818
| 14,771
|
# Hour of Code
Every student in every school should have the opportunity to learn computer science!
Many classrooms have already started the adventure of exposing students to the world of computer science through coding. Some students are making their own video games using Pixel Press Floors and Hopscotch. Others are using the curriculum on code.org and Khan Academy. Scratch, Scratch Jr. and Tynker are other resources to check out if you’re interested in having your kids learn to code. Leave a comment and let us know how your classroom and/or children are learning about computer science!
Make sure to check out the resources on the Hour of Code site at http://csedweek.org.
Also, check out the Computer Science Education Resource Guide @ http://www.smartscholar.com/computer-science-guide/
# Probability
What is the probability of rolling of sum of 7 when rolling 2 dice?
One of my favorite things to do is take a math concept I used to teach my middle school students and see how various younger age groups handle it. More often than not, the instruction can be adjusted so that even the youngest students can learn what some might think is a difficult concept. One of those concepts is probability.
I created a “Dice Experiment” worksheet and conducted it with students in grades 1-6. Kids often associate experiments with science, so it was fun to do a math experiment. We rolled the dice 36 times and recorded the sum. With some groups I gave pairs of students the dice to roll on their own. With other groups, we collected the data as a class.
I loved telling kids that I used to teach this concept to middle school kids. Their faces lit up because they knew they were learning “big kid” math. 🙂
Here is a screenshot of the the experiment worksheet.
With all age groups, we talked about the parts of an experiment and how it’s important to keep track of data. We also talked about tables and charts and how they keep information organized and easy to read. We also talked about patterns and how they’re easier to see when the data is organized. We made lots of predictions and of course, talked about probability. Below you will find a link to a blank copy of the experiment along with an answer key. Feel free to use it and/or tweak it.
Please leave a comment and tell me how you teach probability. Or, give an example of how probability is used in everyday life.
Dice Experiment
Answer Sheet
Dice image taken from http://www.pdclipart.org
# iBook Feedback
Thanks to all the teachers who submitted feedback on our Math Enrichment iBooks. Fifth grade teacher Meredith Chambers from Oakdale was the winner of a \$10 gift card to First Watch. When asked how she is using the Math Enrichment iBooks, Meredith said:
I have a group of five students who passed 5th grade math as 4th graders. I still pretest them to see if they need any reteaching on these topics, but for the most part, after our whole-group lesson, they work with the iBooks for their guided math assignment.
We love getting feedback on the various resources we provide. Please leave a comment on our blog or email us directly with your feedback, input, and suggestions!
# EY Symbaloo
The EY team created a Symbaloo this summer and so far, we’ve made it into all 3rd-6th grade classrooms to show students how to bookmark it on their iPads. We will continue to tweak it as the year goes on and appreciate your feedback and suggestions.
Here is a link with some videos and resources related to the EY Symbaloo: https://ey.westside66.org/wiki/pages/R1x994/Symbaloo.html?target=
# Math Contest Practice grades 1-8
The purpose of Math Stars is to challenge students beyond the classroom setting. Good problems can inspire curiosity about number relationships and geometric properties. It is hoped that in accepting the challenge of mathematical problem solving, students, their parents, and their teachers will be led to explore new mathematical horizons. Math Stars are in sets for Grades 1-8 and include commentaries for teachers. All Math Stars Newsletters are ready for classroom use and available for downloading as PDF files.
Link to newsletters
# Math Problems and QR Codes
Story problems come up in all levels of math and being able to solve problems and explain your thinking is a skill that needs to be practiced over and over again.
This activity provides a way for you to practice your math problem solving skills. Watch the video explanation below and create a math problem of your own.
Link to video explanation
# E/I Participation Form
During E/I time first quarter, many enrichment opportunities have been presented to your students. Feel free to download, print out, or modify the form below to use at conferences. It might be a nice talking point with parents on how students have used their E/I time and what opportunities they have participated in.
Click here for the form
| 1,093
| 4,922
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875
| 4
|
CC-MAIN-2023-06
|
latest
|
en
| 0.935586
|
https://www.reddit.com/r/Economics/comments/da1gg/the_true_cost_of_the_iraq_war_3_trillion_and/
| 1,534,624,763,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221213737.64/warc/CC-MAIN-20180818193409-20180818213409-00055.warc.gz
| 947,359,729
| 27,130
|
×
[–] 7 points8 points (5 children)
Wow, I'm glad we didn't waste that money on European-style universal health care, the upkeep of our own rotting infrastructure, the employment of educators, the development of alternative sources of energy, or any other things that would directly increase the long-term health of the nation.
[–] -3 points-2 points (4 children)
As a libertarian I kept my mouth shut when Bush marched us off to war in 2003. But I got mad - REALLY MAD - when liberals started clamoring for gov't subsidized health care in 2009.
[–] 5 points6 points (1 child)
Sure, keep quiet about the expensive slaughter but woe betide anyone who thinks their taxes should be used to help people.
[–] 2 points3 points (0 children)
He's being sarcastic and misrepresenting the libertarian position, typical smear.
[–] 1 point2 points (0 children)
Why would a libertarian shut their mouth about war? Most libertarians are like Ron Paul in their foreign policy, that is one of non-intervention.
[–] 1 point2 points (0 children)
So war is tolerable , but health care is ? I guess bombs to kill are better than medicines to heal when you really look at it.
[–] 6 points7 points (7 children)
Bad math. It counts the money spent in Iraq at least 3 times that I managed to count.
It adds up the money spent on Iraq, which is proper. It added in the cost to the national deficit. (which is double counting) It also tries to calculate the opportunity cost of the money spent in Iraq, and added in a bunch of costs. Problem is, that is double counting, as spending money already include the opportunity cost.
[–] 3 points4 points (0 children)
To be fair, the rising price of oil is a valid extra cost.
[–] 2 points3 points (2 children)
Problem is, that is double counting, as spending money already include the opportunity cost.
That's not the meaning of opportunity cost - nor does your reading of the "deficit" part appear to be supported by the article.
As an example, suppose you run a small business and a worker embezzles \$100,000.
You're out the \$100K. Since most businesses are run on credit, you're also out the ongoing interest on that extra \$100K of debt you now need to assume - that's the "deficit" part. And if not having the capital makes you lose a job that would have profited you \$10K, that's \$10K opportunity cost. (There are other costs also in the article that could be modelled this way - for example, suppose the employee injured you while stealing, and you need to pay ongoing medical bills...)
While having a Nobel Prize in economics certainly doesn't make you immune to error, and while world-class newspapers certainly do print mistakes, if I were "correcting" a Nobel Prize winner writing about his own field in the Washington Post, I'd certainly make sure my numbers were correct at least twice.
If you want to convince us that you are correct, why not go through the article and extract the specific dollar amounts you believe are double or triple counted? I think if you did that you'd realize that thee is no actual double or triple count, that each dollar only goes into one slot.
[–] 0 points1 point (0 children)
Well, that is a good example, but the federal government don't exactly operate like a small business.
Suppose someone embezzles 100K from the government. The Federal government is out the interest and the 100K, and nothing else. And when you consider how cheap it is for the federal government to borrow money......
The losing the job that profited you 10K doesn't count simply because the US government can always borrow more money with very little interest.
If you want to convince us that you are correct, why not go through the article and extract the specific dollar amounts you believe are double or triple counted? I think if you did that you'd realize that thee is no actual double or triple count, that each dollar only goes into one slot.
Problem is, he didn't actually publish the dollar amounts. He is here is hype a book, and you always need sensationalist numbers to sell books. But if we go though the article and add up proper costs, (medical costs for vets, actual cost of the war, and interest payment calculated from actual interest rates that the government pays) you don't even total up to 1 trillion.
[–] 0 points1 point (0 children)
SMACKDOWN !
[–] 0 points1 point (0 children)
No, that's the true cost. If only they had put it in the headline or something.
[–] -5 points-4 points (1 child)
Liberal Math. Just trying to make a story and please the stupid readers of this rag.
[–] 0 points1 point (0 children)
So speaks the man with a Nazi screen-name.
The last time someone told me I was using Liberal math, it was when I was claiming that the Iraq war might cost as much as \$50 billion, and might even last into 2004.
[–] 0 points1 point (4 children)
[–] 1 point2 points (2 children)
That's a pretty high estimate. Citation?
[–] 3 points4 points (0 children)
http://en.wikipedia.org/wiki/Casualties_of_the_Iraq_War
How many dead people would qualify for an article for you?
[–] 1 point2 points (0 children)
That is an accurate estimation of deaths caused by the US invasion, including those killed directly by US forces or their proxies. It also includes deaths resulting from: the disorder and insecurity the US caused by overthrowing and disbanding the former regime; the lack of power, sanitation, and medication due to US bombardment of civilian infrastructure.
Furthermore, four million refugees have been made refugees. While the US government is celebrating its 'withdrawal' from Iraq, the Iraqis are left to pick up the pieces of their shattered and radicalized country.
[–] 1 point2 points (0 children)
The dollars are salt in the wound. We ended up interfering (once again) in a country we have no business in, and it got the majority of the american people the death of our loved ones and the heavy burden of trillions in debt.
[–] -1 points0 points (2 children)
Clearly the answer is to cease all this socialist spending on VA health care benefits for wounded soldiers.
For that matter, we really ought to cut back on battlefield medicine generally. Thanks to modern technology, many of our troops are surviving injuries that would have been fatal in prior conflicts. This is is simply unacceptable, as many of these people will spend the rest of the lives as invalids with the U.S. tax payer picking of the cost of paying a nurse \$32/hr to wipe some guy's ass for the next five decades. At the very least we ought to look into the possibility of shipping these people off to some place like the Philippines where it costs substantially less to provide that kind of care. Seems like a pareto optimal arrangement.
[–] 0 points1 point (0 children)
You are forgetting about the "Glory". We are a close knit group don't ya know. "Good" people just know what feels right. Never forget!
[–] 0 points1 point (2 children)
freedom comes at a price.
[–] 0 points1 point (0 children)
freedom isnt free. amirite?
[–] 0 points1 point (0 children)
Read in the voice of Buzz Lightyear.
[–] 0 points1 point (2 children)
I didn't realize we were still in a war.
[–] 1 point2 points (0 children)
you aren't, mission accomplished long time ago.
| 1,724
| 7,298
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2018-34
|
latest
|
en
| 0.950229
|
https://math.stackexchange.com/questions/3490652/using-the-limit-comparison-test
| 1,653,793,992,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00290.warc.gz
| 435,883,036
| 65,230
|
# Using the limit comparison test
Given the infinite series: $$\sum^{\infty}_{n=1}\frac{1}{2n+3}$$ Determine whether this series converges.
The answer key used the integral test to determine that no, this series does not converge.
I came at this problem differently. I first tried using the comparison test with $$\frac1n$$ which was inconclusive. I then tried the limit comparison test - again with $$\frac1n$$. I got a limit of $$\frac12$$. Because this is a finite, positive number - the limit diverges.
As a beginner, I am simply unsure that my method was legitimate - after all - its a fifty fifty chance of getting it right:) So, I am asking here- did I find the answer using a legitimate method?
$$\frac{1}{2n + 3} \gt \frac{1}{2n + 4} = \left(\frac{1}{2}\right)\left(\frac{1}{n + 2}\right) \tag{1}\label{eq1A}$$
and $$\sum_{n=1}^{\infty}\frac{1}{n + 2}$$ diverges.
| 262
| 877
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.421875
| 3
|
CC-MAIN-2022-21
|
latest
|
en
| 0.834777
|
https://byjus.com/questions/a-car-with-vertical-windshield-moves-in-a-rainstorm-at-a-speed-of-40-km-hr-the-rain-drops-fall-vertically-with-a-constant-speed-of-20-m-s-the-angle-at-which-raindrops-strike-the-windshield/
| 1,632,649,364,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780057857.27/warc/CC-MAIN-20210926083818-20210926113818-00611.warc.gz
| 209,071,974
| 37,950
|
# A Car With Vertical Windshield Moves In A Rainstorm At A Speed Of 40 Km/Hr. The Rain Drops Fall Vertically With A Constant Speed Of 20 M/S. The Angle At Which Raindrops Strike The Windshield
Speed of car = 40 km/hr.
$\Rightarrow 40 * (\frac{5}{18})$ $\Rightarrow (\frac{100}{9})m/s$ $\Rightarrow tan\Theta =(\frac{v_{c}}{v_{R}})$ $\Rightarrow \Theta = tan=^{-1}(\frac{5}{9})$
Therefore, angle at which raindrops strike the windshield is$\Theta = tan=^{-1}(\frac{5}{9})$
Explore more such questions and answers at BYJU’S.
| 160
| 526
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.921875
| 4
|
CC-MAIN-2021-39
|
latest
|
en
| 0.453357
|
http://www.gradesaver.com/textbooks/science/chemistry/chemistry-the-central-science-13th-edition/chapter-6-electronic-structure-of-atoms-exercises-page-252/6-71b
| 1,519,162,803,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891813109.36/warc/CC-MAIN-20180220204917-20180220224917-00566.warc.gz
| 468,832,673
| 13,050
|
## Chemistry: The Central Science (13th Edition)
The maximum number of electrons that can occupy subshell $5d$ is 10.
*NOTES TO REMEMBER: - The maximum number of electrons which can occupy an orbital is 2. - The number of orbitals in a subshell depends on the type of subshells: +) Subshell $s$: $l=0$, so $m_l=0$. Therefore, 1 orbital. +) Subshell $p$: $l=1$, so $m_l=-1,0,1$. Therefore, 3 orbitals. +) Subshell $d$: $l=2$, so $m_l=-2,-1,0,1,2$. Therefore, 5 orbitals. +) Subshell $f$: $l=3$, so $m_l=-3,-2,-1,0,1,2,3$. Therefore, 7 orbitals. Subshell $5d$ has 5 orbitals. Each orbital can occupy a maximum number of 2 electrons. Therefore, the maximum number of electrons that can occupy subshell $5d$ is 10.
| 242
| 711
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.9375
| 3
|
CC-MAIN-2018-09
|
latest
|
en
| 0.68548
|
https://www.pw.live/question-answer/if--the-value-of-q-will-be:-47902
| 1,669,685,607,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710684.84/warc/CC-MAIN-20221128235805-20221129025805-00471.warc.gz
| 1,014,961,165
| 17,541
|
# . If the value of Q will be:
If then value of Q will be:
A. 4
B. 6
C. 2
D. 5
Solution |||
Value of R = 6 or 5
If R = 6 then P = 2,so Q = 4
If R = 5,then P = 12,which is not possible
So ,Q = 4.
| 95
| 202
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2022-49
|
latest
|
en
| 0.619282
|
http://www.gurufocus.com/term/Intrinsic%20Value%20(DE)/VIAB/Intrinsic%2BValue%2B%2528DE%2529/Viacom%252C%2BInc
| 1,485,236,616,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2017-04/segments/1484560284270.95/warc/CC-MAIN-20170116095124-00396-ip-10-171-10-70.ec2.internal.warc.gz
| 483,351,595
| 29,212
|
Switch to:
GuruFocus has detected 4 Warning Signs with Viacom Inc \$VIAB.
More than 500,000 people have already joined GuruFocus to track the stocks they follow and exchange investment ideas.
Viacom Inc (NAS:VIAB)
Intrinsic Value: DCF (Earnings Based)
\$38.63 (As of Today)
As of today, Viacom Inc's intrinsic value calculated from the Discounted Earnings model is \$38.63.
Note: Discounted Earnings model is only suitable for predictable companies (Business Predictability Rank higher than 1-Star). Result may not be accurate due to the low predictability of business.
Margin of Safety (Earnings Based) using Discounted Earnings model for Viacom Inc is -4.37%.
Definition
This is the intrinsic value calculated from the Discounted Earnings model with default parameters. The calculation method is the same as Discounted Cash Flow model except earnings are used in the calculation instead of free cash flow. This is the default method of calculation with GuruFocus DCF calculator.
Usually a two-stage model is used in calculating the intrinsic value with discounted cash flow model. The first stage is called growth stage; the second is called the terminal stage. In the growth stage the company grows at a faster rate. Because it cannot grow at that rate forever, a lower rate is used for the terminal stage.
GuruFocus DCF calculator is a two-stage model. The default values are defined as:
1. Discount Rate: d=12%
2. Growth Rate in the growth stage: g1=5%
Growth Rate in the growth stage = average earnings growth rate in the past 10 years. If it is higher than 20%, we use 20%. If it is less than 5%, we use 5% instead. => For companies with Average Earnings Growth Rate in the past 10 years less than 5%, GuruFocus defaults => Growth Rate: 5%
3. Years of Growth Stage: y1=10
4. Terminal Growth Rate: g2=4%
5. Years of Terminal Growth: y2=10
6. Earnings Per Share without NRI: eps without nri=\$3.61.
GuruFocus DCF calculator is actually a Discounted Earnings calculator, the Earnings Per Share without NRI is used as the default. The reason we are doing this is we found that historically stock prices are more correlated with earnings than free cash flow.
All of the default settings can be changed and the results are calculated automatically.
Viacom Inc's Intrinsic Value: DCF (Earnings Based) for today is calculated as:
DCF (Earnings Based) = Earnings Per Share * {[(1+g1)/(1+d) + (1+g1)^2/(1+d)^2 + ... + (1+g1)^10/(1+d)^10] + (1+g1)^10/(1+d)^10 * [(1+g2)/(1+d) + (1+g2)^2/(1+d)^2 + ... + (1+g2)^10/(1+d)^10]}
set x = (1+g1)/(1+d) = (1+0.05)/(1+0.12) = 0.9375
and y = (1+g2)/(1+d) = (1+0.04)/(1+0.12) = 0.928571428571
DCF (Earnings Based) = eps * {[x + x^2 + ... + x^10] + x^10 * [y + y^2 + ... + y^10]} = eps * [x * (1-x^10) / (1-x) + x^10 * y * (1-y^10) / (1-y)] = 3.61 * 10.7016 = 38.63
Margin of Safety (Earnings Based) = (DCF (Earnings Based) - Current Price) / DCF (Earnings Based) = (38.632776 - 40.32) / 38.632776 = -4.37 %
* All numbers are in millions except for per share data and ratio. All numbers are in their local exchange's currency.
Explanation
Unlike valuation methods such as Net Current Asset Value, Tangible Book Value per Share, Graham Number, Median Ratio etc, discounted Cash Flow model evaluates the companies based on their future earnings power instead of their assets.
Be Aware
What you need to know about Discounted Earnings model:
1. The Discounted Earnings model evaluates a company based on its future earnings power
2. Growth is taken into account; therefore a faster growth company is worth more if everything else is the same.
3. Since we are projecting future growth, it is assumed that the company will grow at the same rate as it did during the past 10 years. Therefore this model works better for the companies that are relatively consistent performers.
4. The Discounted Earnings model works poorly for inconsistent performers like cyclicals.
5. Your expected return from the investment is a reasonable discount rate assumption.
6. A larger margin of safety should be required for companies with less predictable businesses.
You can screen for stocks that trade below their Intrinsic Value: DCF (FCF Based) and Intrinsic Value: DCF (Earnings Based) with the GuruFocus All-in-One Screener. Companies with a high Predictability Rank that trade at a discount to their Intrinsic Value: DCF (FCF Based) and Intrinsic Value: DCF (Earnings Based) can be found in the screen of Undervalued Predictable Companies.
Related Terms
Get WordPress Plugins for easy affiliate links on Stock Tickers and Guru Names | Earn affiliate commissions by embedding GuruFocus Charts
GuruFocus Affiliate Program: Earn up to \$400 per referral. ( Learn More)
| 1,220
| 4,711
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.59375
| 3
|
CC-MAIN-2017-04
|
longest
|
en
| 0.913455
|
http://mathhelpforum.com/algebra/84240-systems-linear-equations.html
| 1,524,232,014,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125937780.9/warc/CC-MAIN-20180420120351-20180420140351-00442.warc.gz
| 205,050,029
| 11,977
|
# Thread: Systems of Linear Equations
1. ## Systems of Linear Equations
Solve the system of linear equations algebraically.
$\displaystyle x_1 - x_2 + 2x_3 + 2x_4 + 6x_5 = 6$
$\displaystyle 3x_1 - 2x_2 + 4x_3 + 4x_4 + 12x_5 = 14$
$\displaystyle -x_2 - x_3 - x_4 - 3x_5 = -3$
$\displaystyle 2x_1 - 2x_2 + 4x_3 + 5x_4 + 15x_5 = 10$
$\displaystyle 2x_1 - 2x_2 + 4x_3 + 4x_4 + 13x_5 = 13$
I tried to do this many times using various different ways, but I could never get the answer. Help?
I also attached a picture of the original problem for reference.
2. Hello, chrozer!
I used an Augmented Matrix . . . it took me two tries . . .
Solve the system of linear equations algebraically.
. . $\displaystyle \begin{array}{ccc}x_1 - x_2 + 2x_3 + 2x_4 + 6x_5 &=& 6 \\ 3x_1 - 2x_2 + 4x_3 + 4x_4 + 12x_5 &=& 14 \\ \qquad -x_2 - x_3 - x_4 - 3x_5 &=& \text{-}3 \\ 2x_1 - 2x_2 + 4x_3 + 5x_4 + 15x_5 &=& 10 \\ 2x_1 - 2x_2 + 4x_3 + 4x_4 + 13x_5 &=& 13\end{array}$
Multiply the third equation by -1 . . .
We have: .$\displaystyle \left[ \begin{array}{ccccc|c} 1&\text{-}1&2&2&6&6 \\ 3&\text{-}2&4&4&12&14 \\ 0&1&1&1&3&3 \\ 2&\text{-}2&4&5&15&10 \\ 2&\text{-}2&4&4&13&13 \end{array}\right]$
$\displaystyle \begin{array}{c}\\ R_2-3R_1 \\ \\ R_4-2R_1 \\ R_5-R_4 \end{array} \left[ \begin{array}{ccccc|c} 1&\text{-}1&2&2&6&6 \\ 0&1&\text{-}2&\text{-}2&\text{-}6&\text{-}4 \\ 0&1&1&1&3&3 \\ 0&0&0&1&3&\text{-}2 \\ 0&0&0&\text{-}1&\text{-}2&3 \end{array}\right]$
$\displaystyle \begin{array}{c}R_1 +R_2 \\ \\ R_3-R_2 \\ \\ R_5+R_4 \end{array} \left[ \begin{array}{ccccc|c} 1&0&0&0&0&2 \\ 0&1&\text{-}1&\text{-}2&\text{-}6&\text{-}4 \\ 0&0&3&3&9&7 \\ 0&0&0&1&3&\text{-}2 \\ 0&0&0&0&1&5 \end{array}\right]$
. . . $\displaystyle \begin{array}{c} \\ \\ \frac{1}{2}R_3 \\ \\ \\ \end{array} \left[ \begin{array}{ccccc|c} 1&0&0&0&0&2 \\ 0&1&\text{-}2&\text{-}2&\text{-}6&\text{-}4 \\ 0&0&1&1&3&\frac{7}{3} \\ 0&0&0&1&3&\text{-}2 \\ 0&0&0&0&1 & 1 \end{array}\right]$
$\displaystyle \begin{array}{c} \\ R_2+2R_3 \\ R_3 - R_4 \\ R_4 - 3R_5 \\ \\ \end{array}\left[ \begin{array}{ccccc|c} 1&0&0&0&0&2 \\ 0&1&0&0&0&\frac{2}{3} \\ \\[-4mm] 0&0&1&0&0&\frac{13}{3} \\ 0&0&0&1&0& \text{-}5 \\ 0&0&0&0&1&1 \end{array}\right]$
3. Originally Posted by chrozer
[B]I tried to do this many times using various different ways, but I could never get the answer.
Since we cannot see your steps, it will not be possible to find any errors. Sorry!
What methods are you allowed to use? For instance, can you plug this into your graphing calculator? Can you use Cramer's Rule, with your calculator computing the determinants?
Thank you!
4. Originally Posted by stapel
Since we cannot see your steps, it will not be possible to find any errors. Sorry!
What methods are you allowed to use? For instance, can you plug this into your graphing calculator? Can you use Cramer's Rule, with your calculator computing the determinants?
Thank you!
I wish we could use a graphing calculator, it would have made it way easier by using a matrix.
I think we have to solve it algebraically through elimination or substitution. That's what I tried to do as that was the way our teacher thought us.
5. Hello, chrozer!
We can do it by Elimination . . . but it takes longer and is quite messy.
I hate those subscripts; I'll change the variables.
And I'll multiply the third equation by -1.
Solve the system of linear equations algebraically.
$\displaystyle \begin{array}{cccc}a - b + 2c + 2d + 6e &=& 6 & [E1]\\ 3a - 2b + 4c + 4d + 12e &=& 14 &[E2]\\ \qquad b + \;c + \;d + \;3e &=& 3 &[E3]\\ 2a - 2b + 4c + 5d + 15e &=& 10 &[E4]\\ 2a - 2b + 4c + 4d + 13e &=& 13 &[E5]\end{array}$
$\displaystyle \begin{array}{ccccc}\text{-}2(E1)\!: & \text{-}2a + 2b - 4c - 4d - 12e &=&\text{-}12 \\ \text{add }(E4)\!: & 2a - 2b + 4c + 5d + 15e &=& 10 \end{array}$
. . and we have: .$\displaystyle d + 3e \:=\:\text{-}2\;\;[E6]$
$\displaystyle \begin{array}{ccccc}\text{-}1(E4)\!: & \text{-}2a + 2b - 4c - 5d - 15e &=& \text{-}10 \\ \text{add }(E5)\!: & 2a - 2b + 4c + 4c + 13e &=& 13 \end{array}$
. . and we have: .$\displaystyle \text{-}d-2e \:=\:3\;\;[E7]$
Add [E6] and [E7]: .$\displaystyle \boxed{e \:=\:1}$
Substitute into [E6]: .$\displaystyle d + 3(1) \:=\:\text{-}2 \quad\Rightarrow\quad\boxed{d \:=\:\text{-}5}$
Substitute into [E1], [E2], [E3]:
. . $\displaystyle \begin{array}{cccccccc}a-b+2c + 2(\text{-}5) + 6(1) &=& 6 & \Longrightarrow & a - b + 2c &=&10 & [E8] \\ 3a-2b+4c + 4(\text{-}5) + 12(1) &=&14 & \Longrightarrow & 3a - 2b + 4c &=& 22 & [E9] \\ \qquad b+c+(\text{-}5) + 3(1) &=& 3 & \Longrightarrow & \qquad b + c &=& 5 & [E10]\end{array}$
$\displaystyle \begin{array}{cccc}-3(E8)\!:& \text{-}3a + 3b - 6c &=& -30 \\ \text{add }[E9]\!: & 3a-2b + 4c &=& 22 \end{array}$
$\displaystyle \begin{array}{ccccc}\text{We have:} &b - 2c &=& \text{-}8 \\ \text{subtract }(E10)\!: & \text{-}b - c &=& \text{-}5 \end{array}$
And we have: .$\displaystyle \text{-}3c \:=\:\text{-}13 \quad\Rightarrow\quad\boxed{c \:=\:\tfrac{13}{3}}$
Substitute into (E10): .$\displaystyle b + \tfrac{13}{3} \:=\:5 \quad\Rightarrow\quad\boxed{b \:=\:\tfrac{2}{3}}$
Substitute into (E8): .$\displaystyle a - \tfrac{2}{3} + 2\left(\tfrac{13}{3}\right) \:=\:10 \quad\Rightarrow\quad\boxed{a \:=\:2}$
Therefore: .$\displaystyle a \:=\: 2,\quad b\:=\:\tfrac{2}{3},\quad c \:=\: \tfrac{13}{3},\quad d \:=\:\text{-}5, \quad e \:=\: 1$
6. Ok I see what I did wrong now. Thanks for the help.
| 2,398
| 5,430
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.4375
| 4
|
CC-MAIN-2018-17
|
latest
|
en
| 0.389922
|
http://www.gregthatcher.com/Stocks/StockFourierAnalysisDetails.aspx?ticker=LPSN
| 1,529,659,671,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864387.54/warc/CC-MAIN-20180622084714-20180622104714-00147.warc.gz
| 430,520,910
| 96,789
|
Back to list of Stocks See Also: Seasonal Analysis of LPSNGenetic Algorithms Stock Portfolio Generator, and Best Months to Buy/Sell Stocks
# Fourier Analysis of LPSN (LivePerson)
LPSN (LivePerson) appears to have interesting cyclic behaviour every 95 weeks (.6847*cosine), 73 weeks (.4776*sine), and 45 weeks (.3922*cosine).
LPSN (LivePerson) has an average price of 6.9 (topmost row, frequency = 0).
Click on the checkboxes shown on the right to see how the various frequencies contribute to the graph. Look for large magnitude coefficients (sine or cosine), as these are associated with frequencies which contribute most to the associated stock plot. If you find a large magnitude coefficient which dramatically changes the graph, look at the associated "Period" in weeks, as you may have found a significant recurring cycle for the stock of interest.
Right click on the graph above to see the menu of operations (download, full screen, etc.)
## Fourier Analysis
Using data from 4/7/2000 to 6/4/2018 for LPSN (LivePerson), this program was able to calculate the following Fourier Series:
Sequence #Cosine Coefficients Sine Coefficients FrequenciesPeriod
06.90303 0
1-.4493 -4.94737 (1*2π)/949949 weeks
2-1.07197 -.22686 (2*2π)/949475 weeks
32.48857 -.8699 (3*2π)/949316 weeks
4.31633 -1.0063 (4*2π)/949237 weeks
51.64039 -.62456 (5*2π)/949190 weeks
61.02509 -.4151 (6*2π)/949158 weeks
7.05089 -1.10844 (7*2π)/949136 weeks
8-.26407 -.21196 (8*2π)/949119 weeks
9.7187 -.31506 (9*2π)/949105 weeks
10.68474 -.19469 (10*2π)/94995 weeks
11-.09432 -.08774 (11*2π)/94986 weeks
12.25044 .02028 (12*2π)/94979 weeks
13.2658 -.47764 (13*2π)/94973 weeks
14-.34106 -.12956 (14*2π)/94968 weeks
15.37249 .07315 (15*2π)/94963 weeks
16.08354 .0748 (16*2π)/94959 weeks
17-.07804 .09426 (17*2π)/94956 weeks
18.0228 -.08711 (18*2π)/94953 weeks
19.10808 .19984 (19*2π)/94950 weeks
20.38577 .2417 (20*2π)/94947 weeks
21.39222 -.22128 (21*2π)/94945 weeks
22.05191 -.27613 (22*2π)/94943 weeks
23-.21976 -.08231 (23*2π)/94941 weeks
24.28254 .06369 (24*2π)/94940 weeks
25.14935 -.28084 (25*2π)/94938 weeks
26.0081 -.29566 (26*2π)/94937 weeks
27-.15848 -.1567 (27*2π)/94935 weeks
28-.02214 -.10955 (28*2π)/94934 weeks
29-.03493 -.1854 (29*2π)/94933 weeks
30-.1694 -.26341 (30*2π)/94932 weeks
31.01553 -.14966 (31*2π)/94931 weeks
32-.17688 -.18621 (32*2π)/94930 weeks
33-.10746 -.11358 (33*2π)/94929 weeks
34.05964 -.10718 (34*2π)/94928 weeks
35-.10206 -.25047 (35*2π)/94927 weeks
36-.00334 -.17786 (36*2π)/94926 weeks
37-.0451 -.00765 (37*2π)/94926 weeks
38.00995 -.25295 (38*2π)/94925 weeks
39-.03316 -.31124 (39*2π)/94924 weeks
40-.15082 -.06286 (40*2π)/94924 weeks
41.09483 -.082 (41*2π)/94923 weeks
42.07853 -.12833 (42*2π)/94923 weeks
43-.12618 -.12356 (43*2π)/94922 weeks
44.00916 -.07713 (44*2π)/94922 weeks
45.1546 -.18989 (45*2π)/94921 weeks
46-.05716 -.04508 (46*2π)/94921 weeks
47-.06673 -.07303 (47*2π)/94920 weeks
48-.00883 -.01777 (48*2π)/94920 weeks
49.08105 -.06518 (49*2π)/94919 weeks
50.0275 -.0394 (50*2π)/94919 weeks
51.038 -.01785 (51*2π)/94919 weeks
52.06675 -.02816 (52*2π)/94918 weeks
53.04555 -.16181 (53*2π)/94918 weeks
54-.02676 -.08365 (54*2π)/94918 weeks
55.05091 .01385 (55*2π)/94917 weeks
56.05528 -.14228 (56*2π)/94917 weeks
57-.02072 -.05428 (57*2π)/94917 weeks
58.12311 -.1513 (58*2π)/94916 weeks
59-.0881 -.15241 (59*2π)/94916 weeks
60-.01653 .01584 (60*2π)/94916 weeks
61.08724 -.08931 (61*2π)/94916 weeks
62-.09175 -.10761 (62*2π)/94915 weeks
63-.02019 -.00297 (63*2π)/94915 weeks
64.03715 -.12946 (64*2π)/94915 weeks
65-.08922 -.06264 (65*2π)/94915 weeks
66.02489 .069 (66*2π)/94914 weeks
67.04849 -.06529 (67*2π)/94914 weeks
68-.02822 -.09391 (68*2π)/94914 weeks
69-.0153 -.07762 (69*2π)/94914 weeks
70.01161 .01368 (70*2π)/94914 weeks
71.01733 -.06008 (71*2π)/94913 weeks
72-.09329 -.08141 (72*2π)/94913 weeks
73.15199 .0539 (73*2π)/94913 weeks
74.00019 -.15776 (74*2π)/94913 weeks
75-.06961 -.06274 (75*2π)/94913 weeks
76.06168 -.05305 (76*2π)/94912 weeks
77-.00897 -.08619 (77*2π)/94912 weeks
78.03467 -.02393 (78*2π)/94912 weeks
79.00805 -.06204 (79*2π)/94912 weeks
80.00049 -.09482 (80*2π)/94912 weeks
81.06312 -.07255 (81*2π)/94912 weeks
82.06999 -.05779 (82*2π)/94912 weeks
83-.01353 -.01361 (83*2π)/94911 weeks
84.03259 -.10777 (84*2π)/94911 weeks
85.03491 -.08654 (85*2π)/94911 weeks
86.07415 -.07192 (86*2π)/94911 weeks
87-.02072 -.08777 (87*2π)/94911 weeks
88-.01574 -.01402 (88*2π)/94911 weeks
89.02757 -.05098 (89*2π)/94911 weeks
90.00843 -.12307 (90*2π)/94911 weeks
91.0456 -.08084 (91*2π)/94910 weeks
92-.05964 -.06564 (92*2π)/94910 weeks
93-.07734 -.07134 (93*2π)/94910 weeks
94-.00533 -.0266 (94*2π)/94910 weeks
95-.03987 -.05021 (95*2π)/94910 weeks
96-.04172 -.08188 (96*2π)/94910 weeks
97-.08558 -.03594 (97*2π)/94910 weeks
98-.08263 .02132 (98*2π)/94910 weeks
99.022 -.04565 (99*2π)/94910 weeks
100-.0428 -.05069 (100*2π)/9499 weeks
101-.06226 -.01221 (101*2π)/9499 weeks
102-.01914 -.02349 (102*2π)/9499 weeks
103-.07622 -.02319 (103*2π)/9499 weeks
104.01728 .02406 (104*2π)/9499 weeks
105.01758 -.05391 (105*2π)/9499 weeks
106-.04106 -.04077 (106*2π)/9499 weeks
107-.00132 -.01281 (107*2π)/9499 weeks
108-.06766 -.10116 (108*2π)/9499 weeks
109-.00497 .00236 (109*2π)/9499 weeks
110-.00792 -.03525 (110*2π)/9499 weeks
111-.00072 -.05896 (111*2π)/9499 weeks
112-.02295 -.06229 (112*2π)/9498 weeks
113-.05626 -.03933 (113*2π)/9498 weeks
114.0071 .0027 (114*2π)/9498 weeks
115.00778 -.04991 (115*2π)/9498 weeks
116.01943 -.07469 (116*2π)/9498 weeks
117-.02226 -.07427 (117*2π)/9498 weeks
118-.00135 -.02228 (118*2π)/9498 weeks
119-.01937 -.0428 (119*2π)/9498 weeks
120.02663 -.00843 (120*2π)/9498 weeks
121.02693 -.08946 (121*2π)/9498 weeks
122-.03753 -.03229 (122*2π)/9498 weeks
123.04059 .03905 (123*2π)/9498 weeks
124.01169 -.0596 (124*2π)/9498 weeks
125-.01591 -.03836 (125*2π)/9498 weeks
126.00219 .00865 (126*2π)/9498 weeks
127.07466 .00386 (127*2π)/9497 weeks
128-.02587 -.03425 (128*2π)/9497 weeks
129.00339 -.00139 (129*2π)/9497 weeks
130.0255 -.01203 (130*2π)/9497 weeks
131.0084 .02017 (131*2π)/9497 weeks
132.0515 -.0466 (132*2π)/9497 weeks
133-.02205 -.00953 (133*2π)/9497 weeks
134.01994 .02083 (134*2π)/9497 weeks
135.00136 -.0528 (135*2π)/9497 weeks
136.0222 .00545 (136*2π)/9497 weeks
137-.00057 -.00942 (137*2π)/9497 weeks
138.00418 -.01449 (138*2π)/9497 weeks
139-.00211 -.03792 (139*2π)/9497 weeks
140-.00627 -.01683 (140*2π)/9497 weeks
141-.00737 -.02144 (141*2π)/9497 weeks
142-.00617 .00071 (142*2π)/9497 weeks
143.0164 -.0522 (143*2π)/9497 weeks
144-.0111 -.02432 (144*2π)/9497 weeks
145.00462 -.03483 (145*2π)/9497 weeks
146-.01439 -.05021 (146*2π)/9497 weeks
147-.04488 -.0328 (147*2π)/9496 weeks
148-.01392 -.04214 (148*2π)/9496 weeks
149-.02213 -.06453 (149*2π)/9496 weeks
150-.04339 -.01061 (150*2π)/9496 weeks
151.00958 -.01342 (151*2π)/9496 weeks
152.01997 -.07259 (152*2π)/9496 weeks
153-.05081 -.06749 (153*2π)/9496 weeks
154-.01625 -.04528 (154*2π)/9496 weeks
155-.02706 -.04201 (155*2π)/9496 weeks
156-.01499 -.02514 (156*2π)/9496 weeks
157-.01961 -.04281 (157*2π)/9496 weeks
158-.03764 -.02825 (158*2π)/9496 weeks
159.00353 .00427 (159*2π)/9496 weeks
160-.00011 -.03078 (160*2π)/9496 weeks
161-.01585 .01167 (161*2π)/9496 weeks
162.02006 -.00907 (162*2π)/9496 weeks
163.00702 -.03466 (163*2π)/9496 weeks
164-.02418 -.02064 (164*2π)/9496 weeks
165.03839 .00683 (165*2π)/9496 weeks
166-.02061 -.01394 (166*2π)/9496 weeks
167.01043 .00247 (167*2π)/9496 weeks
168.02065 -.02354 (168*2π)/9496 weeks
169-.01895 -.00493 (169*2π)/9496 weeks
170.07204 -.00953 (170*2π)/9496 weeks
171-.00421 -.04886 (171*2π)/9496 weeks
172-.01834 -.0312 (172*2π)/9496 weeks
173.00084 -.01607 (173*2π)/9495 weeks
174-.03197 -.03313 (174*2π)/9495 weeks
175.0194 -.00178 (175*2π)/9495 weeks
176-.0002 -.02769 (176*2π)/9495 weeks
177-.0323 -.00187 (177*2π)/9495 weeks
178.0091 .007 (178*2π)/9495 weeks
179-.01411 -.01804 (179*2π)/9495 weeks
180.01259 -.00542 (180*2π)/9495 weeks
181.01165 -.0219 (181*2π)/9495 weeks
182-.01712 -.00402 (182*2π)/9495 weeks
183-.00692 .01233 (183*2π)/9495 weeks
184.04325 -.02252 (184*2π)/9495 weeks
185-.01053 -.04989 (185*2π)/9495 weeks
186-.02496 -.00283 (186*2π)/9495 weeks
187.00518 -.0017 (187*2π)/9495 weeks
188-.00286 -.02348 (188*2π)/9495 weeks
189-.00566 -.03336 (189*2π)/9495 weeks
190-.01216 -.01575 (190*2π)/9495 weeks
191.00321 -.01305 (191*2π)/9495 weeks
192-.00446 -.03716 (192*2π)/9495 weeks
193.00368 -.00076 (193*2π)/9495 weeks
194-.01666 -.04093 (194*2π)/9495 weeks
195-.03301 -.01245 (195*2π)/9495 weeks
196.01343 -.02208 (196*2π)/9495 weeks
197-.02743 -.0121 (197*2π)/9495 weeks
198.00243 .00363 (198*2π)/9495 weeks
199.01654 -.04596 (199*2π)/9495 weeks
200-.0197 -.04207 (200*2π)/9495 weeks
201-.02386 -.00924 (201*2π)/9495 weeks
202-.01149 -.00597 (202*2π)/9495 weeks
203-.00448 -.01329 (203*2π)/9495 weeks
204-.01364 -.04994 (204*2π)/9495 weeks
205-.02198 -.02437 (205*2π)/9495 weeks
206-.01327 .01289 (206*2π)/9495 weeks
207-.0132 -.02976 (207*2π)/9495 weeks
208-.01734 -.04341 (208*2π)/9495 weeks
209-.03146 -.00552 (209*2π)/9495 weeks
210-.00023 -.03401 (210*2π)/9495 weeks
211-.03804 -.00075 (211*2π)/9494 weeks
212-.01653 -.0076 (212*2π)/9494 weeks
213-.02556 -.04769 (213*2π)/9494 weeks
214-.0152 -.01975 (214*2π)/9494 weeks
215-.02188 .0072 (215*2π)/9494 weeks
216.00712 -.0007 (216*2π)/9494 weeks
217-.00846 -.03088 (217*2π)/9494 weeks
218-.0273 -.01985 (218*2π)/9494 weeks
219.00887 -.01595 (219*2π)/9494 weeks
220.01571 -.0003 (220*2π)/9494 weeks
221-.0087 -.0167 (221*2π)/9494 weeks
222-.01906 -.03542 (222*2π)/9494 weeks
223-.00192 -.02722 (223*2π)/9494 weeks
224-.00249 -.02068 (224*2π)/9494 weeks
225-.01261 -.02986 (225*2π)/9494 weeks
226-.02113 .02116 (226*2π)/9494 weeks
227-.00804 -.01234 (227*2π)/9494 weeks
228-.00441 -.02965 (228*2π)/9494 weeks
229-.00417 .00499 (229*2π)/9494 weeks
230.00743 .00324 (230*2π)/9494 weeks
231.00017 -.01997 (231*2π)/9494 weeks
232-.02347 -.01444 (232*2π)/9494 weeks
233.01851 -.00404 (233*2π)/9494 weeks
234-.01729 -.02712 (234*2π)/9494 weeks
235-.00799 -.00716 (235*2π)/9494 weeks
236-.01235 -.00213 (236*2π)/9494 weeks
237.01151 -.01065 (237*2π)/9494 weeks
238-.00019 -.03979 (238*2π)/9494 weeks
239-.05588 -.01425 (239*2π)/9494 weeks
240-.00987 -.00409 (240*2π)/9494 weeks
241-.01827 -.01149 (241*2π)/9494 weeks
242-.01476 .00627 (242*2π)/9494 weeks
243-.01048 -.01715 (243*2π)/9494 weeks
244-.0498 .01557 (244*2π)/9494 weeks
245.02172 .00708 (245*2π)/9494 weeks
246-.00471 -.03304 (246*2π)/9494 weeks
247-.04285 .01037 (247*2π)/9494 weeks
248.00993 .00483 (248*2π)/9494 weeks
249-.01184 -.015 (249*2π)/9494 weeks
250.00594 -.01617 (250*2π)/9494 weeks
251-.00376 .00583 (251*2π)/9494 weeks
252-.00094 -.00488 (252*2π)/9494 weeks
253.0061 -.02003 (253*2π)/9494 weeks
254-.00638 -.02593 (254*2π)/9494 weeks
255-.00144 -.0226 (255*2π)/9494 weeks
256-.00877 -.0131 (256*2π)/9494 weeks
257-.00558 -.0186 (257*2π)/9494 weeks
258-.00596 -.018 (258*2π)/9494 weeks
259-.00307 -.01866 (259*2π)/9494 weeks
260-.01079 -.06401 (260*2π)/9494 weeks
261-.05426 .01718 (261*2π)/9494 weeks
262.01129 .0113 (262*2π)/9494 weeks
263-.00918 -.02589 (263*2π)/9494 weeks
264-.01717 -.01004 (264*2π)/9494 weeks
265.00117 -.03362 (265*2π)/9494 weeks
266-.00386 -.00766 (266*2π)/9494 weeks
267-.01456 -.01278 (267*2π)/9494 weeks
268-.01233 -.01927 (268*2π)/9494 weeks
269-.00067 -.0229 (269*2π)/9494 weeks
270-.00158 -.03273 (270*2π)/9494 weeks
271-.03375 -.02389 (271*2π)/9494 weeks
272-.02399 .01708 (272*2π)/9493 weeks
273-.00349 -.03953 (273*2π)/9493 weeks
274-.02417 -.01053 (274*2π)/9493 weeks
275-.01308 .00526 (275*2π)/9493 weeks
276-.00647 -.02147 (276*2π)/9493 weeks
277-.00902 .00807 (277*2π)/9493 weeks
278-.01632 -.0094 (278*2π)/9493 weeks
279-.01922 -.01424 (279*2π)/9493 weeks
280-.02351 .01671 (280*2π)/9493 weeks
281-.00243 .00887 (281*2π)/9493 weeks
282-.00423 -.03257 (282*2π)/9493 weeks
283-.03997 .01743 (283*2π)/9493 weeks
284-.00884 .02489 (284*2π)/9493 weeks
285.01657 -.01419 (285*2π)/9493 weeks
286-.02159 -.01451 (286*2π)/9493 weeks
287-.01031 .02461 (287*2π)/9493 weeks
288.00456 -.01644 (288*2π)/9493 weeks
289-.02677 -.01292 (289*2π)/9493 weeks
290-.01808 .00104 (290*2π)/9493 weeks
291-.01773 -.0009 (291*2π)/9493 weeks
292-.00582 -.00245 (292*2π)/9493 weeks
293-.00989 -.00907 (293*2π)/9493 weeks
294-.01497 -.01255 (294*2π)/9493 weeks
295-.02119 -.01038 (295*2π)/9493 weeks
296-.0108 -.00756 (296*2π)/9493 weeks
297-.02212 -.01413 (297*2π)/9493 weeks
298-.01367 -.00701 (298*2π)/9493 weeks
299-.01382 -.0125 (299*2π)/9493 weeks
300-.02887 -.01898 (300*2π)/9493 weeks
301-.02121 -.01813 (301*2π)/9493 weeks
302.00375 -.02486 (302*2π)/9493 weeks
303-.02922 -.02687 (303*2π)/9493 weeks
304-.03863 -.0074 (304*2π)/9493 weeks
305-.00843 -.01055 (305*2π)/9493 weeks
306-.01968 -.03291 (306*2π)/9493 weeks
307-.05444 -.02956 (307*2π)/9493 weeks
308-.0194 .0116 (308*2π)/9493 weeks
309-.01999 -.03253 (309*2π)/9493 weeks
310-.04357 -.0046 (310*2π)/9493 weeks
311-.00993 .00759 (311*2π)/9493 weeks
312-.02948 -.01008 (312*2π)/9493 weeks
313-.00909 -.00506 (313*2π)/9493 weeks
314-.00391 -.01035 (314*2π)/9493 weeks
315-.01929 .00399 (315*2π)/9493 weeks
316.00025 -.0018 (316*2π)/9493 weeks
317-.00903 -.03343 (317*2π)/9493 weeks
318-.02261 -.02724 (318*2π)/9493 weeks
319-.01537 -.00956 (319*2π)/9493 weeks
320-.01804 .00404 (320*2π)/9493 weeks
321-.00278 -.01783 (321*2π)/9493 weeks
322-.02159 -.02392 (322*2π)/9493 weeks
323-.00282 .00144 (323*2π)/9493 weeks
324-.00581 -.02595 (324*2π)/9493 weeks
325-.00488 -.01961 (325*2π)/9493 weeks
326-.0114 .00386 (326*2π)/9493 weeks
327-.01078 -.0295 (327*2π)/9493 weeks
328-.02207 -.01484 (328*2π)/9493 weeks
329-.00234 .00732 (329*2π)/9493 weeks
330.01461 -.01045 (330*2π)/9493 weeks
331-.01942 -.00691 (331*2π)/9493 weeks
332-.01215 .00293 (332*2π)/9493 weeks
333.00563 .00611 (333*2π)/9493 weeks
334.01822 -.00822 (334*2π)/9493 weeks
335-.00756 -.01006 (335*2π)/9493 weeks
336-.01709 .00741 (336*2π)/9493 weeks
337.02165 -.00282 (337*2π)/9493 weeks
338-.01476 -.02287 (338*2π)/9493 weeks
339-.01309 .00153 (339*2π)/9493 weeks
340.00109 .00684 (340*2π)/9493 weeks
341-.00659 -.02049 (341*2π)/9493 weeks
342-.02165 .00543 (342*2π)/9493 weeks
343-.01338 -.00312 (343*2π)/9493 weeks
344-.00973 -.00366 (344*2π)/9493 weeks
345-.00809 .02611 (345*2π)/9493 weeks
346.00212 -.02578 (346*2π)/9493 weeks
347-.02739 .01557 (347*2π)/9493 weeks
348.0067 .01415 (348*2π)/9493 weeks
349.01103 -.03282 (349*2π)/9493 weeks
350-.02703 .00871 (350*2π)/9493 weeks
351.01214 .0067 (351*2π)/9493 weeks
352-.01371 -.03923 (352*2π)/9493 weeks
353-.02921 -.01094 (353*2π)/9493 weeks
354-.0009 -.00875 (354*2π)/9493 weeks
355-.00191 -.01304 (355*2π)/9493 weeks
356-.03633 -.02566 (356*2π)/9493 weeks
357-.0254 -.01395 (357*2π)/9493 weeks
358-.01969 -.0296 (358*2π)/9493 weeks
359-.03178 -.02089 (359*2π)/9493 weeks
360-.02474 -.00066 (360*2π)/9493 weeks
361-.0333 .00276 (361*2π)/9493 weeks
362-.01227 -.00523 (362*2π)/9493 weeks
363-.02189 -.03081 (363*2π)/9493 weeks
364-.01324 -.00004 (364*2π)/9493 weeks
365-.01443 -.01322 (365*2π)/9493 weeks
366-.01728 .0043 (366*2π)/9493 weeks
367-.00646 -.01966 (367*2π)/9493 weeks
368-.02266 -.01686 (368*2π)/9493 weeks
369-.00513 -.00892 (369*2π)/9493 weeks
370-.00238 -.01942 (370*2π)/9493 weeks
371-.00528 -.00315 (371*2π)/9493 weeks
372-.00966 -.02018 (372*2π)/9493 weeks
373.00161 -.03226 (373*2π)/9493 weeks
374-.0274 -.02425 (374*2π)/9493 weeks
375-.0032 .00421 (375*2π)/9493 weeks
376.00523 -.01954 (376*2π)/9493 weeks
377-.01486 -.03227 (377*2π)/9493 weeks
378-.02768 -.03891 (378*2π)/9493 weeks
379-.02939 .00092 (379*2π)/9493 weeks
380.00147 -.00943 (380*2π)/9492 weeks
381-.02799 -.01693 (381*2π)/9492 weeks
382-.02547 -.00888 (382*2π)/9492 weeks
383-.02367 -.014 (383*2π)/9492 weeks
384-.02157 .00268 (384*2π)/9492 weeks
385-.01237 .00078 (385*2π)/9492 weeks
386-.02281 -.0161 (386*2π)/9492 weeks
387-.01167 .01991 (387*2π)/9492 weeks
388-.0074 -.02804 (388*2π)/9492 weeks
389-.03523 -.01145 (389*2π)/9492 weeks
390-.01621 .02629 (390*2π)/9492 weeks
391-.00354 -.01756 (391*2π)/9492 weeks
392-.03242 -.00414 (392*2π)/9492 weeks
393-.01504 .01399 (393*2π)/9492 weeks
394-.02075 .00424 (394*2π)/9492 weeks
395-.00311 .00462 (395*2π)/9492 weeks
396.0001 -.0222 (396*2π)/9492 weeks
397-.02634 .00743 (397*2π)/9492 weeks
398-.01322 -.0112 (398*2π)/9492 weeks
399-.03204 -.01176 (399*2π)/9492 weeks
400-.01552 -.00266 (400*2π)/9492 weeks
401-.01544 .00144 (401*2π)/9492 weeks
402-.01374 -.00625 (402*2π)/9492 weeks
403-.03034 -.01811 (403*2π)/9492 weeks
404-.03814 -.01409 (404*2π)/9492 weeks
405-.00857 -.0014 (405*2π)/9492 weeks
406-.0201 -.02256 (406*2π)/9492 weeks
407-.03421 .00651 (407*2π)/9492 weeks
408-.00488 -.00001 (408*2π)/9492 weeks
409-.01361 -.03035 (409*2π)/9492 weeks
410-.01858 -.01734 (410*2π)/9492 weeks
411-.02003 .01218 (411*2π)/9492 weeks
412.0097 -.01512 (412*2π)/9492 weeks
413-.02976 -.03058 (413*2π)/9492 weeks
414-.01663 .00789 (414*2π)/9492 weeks
415.00082 -.01872 (415*2π)/9492 weeks
416-.0152 -.01539 (416*2π)/9492 weeks
417-.00808 .00715 (417*2π)/9492 weeks
418-.00651 -.01656 (418*2π)/9492 weeks
419-.02464 -.00215 (419*2π)/9492 weeks
420.00762 .00177 (420*2π)/9492 weeks
421-.00472 .00479 (421*2π)/9492 weeks
422-.0171 .00068 (422*2π)/9492 weeks
423.02065 .00227 (423*2π)/9492 weeks
424-.00495 -.02312 (424*2π)/9492 weeks
425-.03128 -.01193 (425*2π)/9492 weeks
426.00825 .01327 (426*2π)/9492 weeks
427-.01047 -.00046 (427*2π)/9492 weeks
428-.03341 .00074 (428*2π)/9492 weeks
429.00166 .00023 (429*2π)/9492 weeks
430-.00234 .00102 (430*2π)/9492 weeks
431.00267 .0051 (431*2π)/9492 weeks
432-.01063 .00031 (432*2π)/9492 weeks
433-.00199 .00992 (433*2π)/9492 weeks
434.00191 -.02382 (434*2π)/9492 weeks
435-.01041 .00487 (435*2π)/9492 weeks
436.00085 -.0098 (436*2π)/9492 weeks
437-.00249 -.00267 (437*2π)/9492 weeks
438-.00863 -.01123 (438*2π)/9492 weeks
439-.0089 -.01838 (439*2π)/9492 weeks
440-.01032 -.0139 (440*2π)/9492 weeks
441-.01037 -.0007 (441*2π)/9492 weeks
442.01913 .01074 (442*2π)/9492 weeks
443-.01921 -.03422 (443*2π)/9492 weeks
444-.01631 -.00526 (444*2π)/9492 weeks
445-.00323 -.00944 (445*2π)/9492 weeks
446-.00319 -.00788 (446*2π)/9492 weeks
447.00521 .00061 (447*2π)/9492 weeks
448-.00478 -.01942 (448*2π)/9492 weeks
449-.01744 -.02046 (449*2π)/9492 weeks
450-.00305 -.00576 (450*2π)/9492 weeks
451.00557 -.00226 (451*2π)/9492 weeks
452-.00277 -.03035 (452*2π)/9492 weeks
453-.04621 -.01613 (453*2π)/9492 weeks
454.0005 .01919 (454*2π)/9492 weeks
455.00057 -.01146 (455*2π)/9492 weeks
456.00564 -.0111 (456*2π)/9492 weeks
457-.01809 -.02093 (457*2π)/9492 weeks
458-.01568 .00439 (458*2π)/9492 weeks
459-.01165 -.01119 (459*2π)/9492 weeks
460-.0082 .01099 (460*2π)/9492 weeks
461-.00307 .0006 (461*2π)/9492 weeks
462-.01797 -.0035 (462*2π)/9492 weeks
463.00795 .00294 (463*2π)/9492 weeks
464-.00526 -.0056 (464*2π)/9492 weeks
465-.01654 -.01149 (465*2π)/9492 weeks
466-.0083 .0197 (466*2π)/9492 weeks
467.00759 .00923 (467*2π)/9492 weeks
468-.00691 -.00286 (468*2π)/9492 weeks
469.02339 -.00497 (469*2π)/9492 weeks
470-.01655 -.01092 (470*2π)/9492 weeks
471-.0264 -.01511 (471*2π)/9492 weeks
472-.00518 .00911 (472*2π)/9492 weeks
473-.00505 -.00452 (473*2π)/9492 weeks
474-.01858 -.00723 (474*2π)/9492 weeks
475-.01858 .00723 (475*2π)/9492 weeks
476-.00505 .00452 (476*2π)/9492 weeks
477-.00518 -.00911 (477*2π)/9492 weeks
478-.0264 .01511 (478*2π)/9492 weeks
479-.01655 .01092 (479*2π)/9492 weeks
480.02339 .00497 (480*2π)/9492 weeks
481-.00691 .00286 (481*2π)/9492 weeks
482.00759 -.00923 (482*2π)/9492 weeks
483-.0083 -.0197 (483*2π)/9492 weeks
484-.01654 .01149 (484*2π)/9492 weeks
485-.00526 .0056 (485*2π)/9492 weeks
486.00795 -.00294 (486*2π)/9492 weeks
487-.01797 .0035 (487*2π)/9492 weeks
488-.00307 -.0006 (488*2π)/9492 weeks
489-.0082 -.01099 (489*2π)/9492 weeks
490-.01165 .01119 (490*2π)/9492 weeks
491-.01568 -.00439 (491*2π)/9492 weeks
492-.01809 .02093 (492*2π)/9492 weeks
493.00564 .0111 (493*2π)/9492 weeks
494.00057 .01146 (494*2π)/9492 weeks
495.0005 -.01919 (495*2π)/9492 weeks
496-.04621 .01613 (496*2π)/9492 weeks
497-.00277 .03035 (497*2π)/9492 weeks
498.00557 .00226 (498*2π)/9492 weeks
499-.00305 .00576 (499*2π)/9492 weeks
500-.01744 .02046 (500*2π)/9492 weeks
501-.00478 .01942 (501*2π)/9492 weeks
502.00521 -.00061 (502*2π)/9492 weeks
503-.00319 .00788 (503*2π)/9492 weeks
504-.00323 .00944 (504*2π)/9492 weeks
505-.01631 .00526 (505*2π)/9492 weeks
506-.01921 .03422 (506*2π)/9492 weeks
507.01913 -.01074 (507*2π)/9492 weeks
508-.01037 .0007 (508*2π)/9492 weeks
509-.01032 .0139 (509*2π)/9492 weeks
510-.0089 .01838 (510*2π)/9492 weeks
511-.00863 .01123 (511*2π)/9492 weeks
512-.00249 .00267 (512*2π)/9492 weeks
513.00085 .0098 (513*2π)/9492 weeks
514-.01041 -.00487 (514*2π)/9492 weeks
515.00191 .02382 (515*2π)/9492 weeks
516-.00199 -.00992 (516*2π)/9492 weeks
517-.01063 -.00031 (517*2π)/9492 weeks
518.00267 -.0051 (518*2π)/9492 weeks
519-.00234 -.00102 (519*2π)/9492 weeks
520.00166 -.00023 (520*2π)/9492 weeks
521-.03341 -.00074 (521*2π)/9492 weeks
522-.01047 .00046 (522*2π)/9492 weeks
523.00825 -.01327 (523*2π)/9492 weeks
524-.03128 .01193 (524*2π)/9492 weeks
525-.00495 .02312 (525*2π)/9492 weeks
526.02065 -.00227 (526*2π)/9492 weeks
527-.0171 -.00068 (527*2π)/9492 weeks
528-.00472 -.00479 (528*2π)/9492 weeks
529.00762 -.00177 (529*2π)/9492 weeks
530-.02464 .00215 (530*2π)/9492 weeks
531-.00651 .01656 (531*2π)/9492 weeks
532-.00808 -.00715 (532*2π)/9492 weeks
533-.0152 .01539 (533*2π)/9492 weeks
534.00082 .01872 (534*2π)/9492 weeks
535-.01663 -.00789 (535*2π)/9492 weeks
536-.02976 .03058 (536*2π)/9492 weeks
537.0097 .01512 (537*2π)/9492 weeks
538-.02003 -.01218 (538*2π)/9492 weeks
539-.01858 .01734 (539*2π)/9492 weeks
540-.01361 .03035 (540*2π)/9492 weeks
541-.00488 .00001 (541*2π)/9492 weeks
542-.03421 -.00651 (542*2π)/9492 weeks
543-.0201 .02256 (543*2π)/9492 weeks
544-.00857 .0014 (544*2π)/9492 weeks
545-.03814 .01409 (545*2π)/9492 weeks
546-.03034 .01811 (546*2π)/9492 weeks
547-.01374 .00625 (547*2π)/9492 weeks
548-.01544 -.00144 (548*2π)/9492 weeks
549-.01552 .00266 (549*2π)/9492 weeks
550-.03204 .01176 (550*2π)/9492 weeks
551-.01322 .0112 (551*2π)/9492 weeks
552-.02634 -.00743 (552*2π)/9492 weeks
553.0001 .0222 (553*2π)/9492 weeks
554-.00311 -.00462 (554*2π)/9492 weeks
555-.02075 -.00424 (555*2π)/9492 weeks
556-.01504 -.01399 (556*2π)/9492 weeks
557-.03242 .00414 (557*2π)/9492 weeks
558-.00354 .01756 (558*2π)/9492 weeks
559-.01621 -.02629 (559*2π)/9492 weeks
560-.03523 .01145 (560*2π)/9492 weeks
561-.0074 .02804 (561*2π)/9492 weeks
562-.01167 -.01991 (562*2π)/9492 weeks
563-.02281 .0161 (563*2π)/9492 weeks
564-.01237 -.00078 (564*2π)/9492 weeks
565-.02157 -.00268 (565*2π)/9492 weeks
566-.02367 .014 (566*2π)/9492 weeks
567-.02547 .00888 (567*2π)/9492 weeks
568-.02799 .01693 (568*2π)/9492 weeks
569.00147 .00943 (569*2π)/9492 weeks
570-.02939 -.00092 (570*2π)/9492 weeks
571-.02768 .03891 (571*2π)/9492 weeks
572-.01486 .03227 (572*2π)/9492 weeks
573.00523 .01954 (573*2π)/9492 weeks
574-.0032 -.00421 (574*2π)/9492 weeks
575-.0274 .02425 (575*2π)/9492 weeks
576.00161 .03226 (576*2π)/9492 weeks
577-.00966 .02018 (577*2π)/9492 weeks
578-.00528 .00315 (578*2π)/9492 weeks
579-.00238 .01942 (579*2π)/9492 weeks
580-.00513 .00892 (580*2π)/9492 weeks
581-.02266 .01686 (581*2π)/9492 weeks
582-.00646 .01966 (582*2π)/9492 weeks
583-.01728 -.0043 (583*2π)/9492 weeks
584-.01443 .01322 (584*2π)/9492 weeks
585-.01324 .00004 (585*2π)/9492 weeks
586-.02189 .03081 (586*2π)/9492 weeks
587-.01227 .00523 (587*2π)/9492 weeks
588-.0333 -.00276 (588*2π)/9492 weeks
589-.02474 .00066 (589*2π)/9492 weeks
590-.03178 .02089 (590*2π)/9492 weeks
591-.01969 .0296 (591*2π)/9492 weeks
592-.0254 .01395 (592*2π)/9492 weeks
593-.03633 .02566 (593*2π)/9492 weeks
594-.00191 .01304 (594*2π)/9492 weeks
595-.0009 .00875 (595*2π)/9492 weeks
596-.02921 .01094 (596*2π)/9492 weeks
597-.01371 .03923 (597*2π)/9492 weeks
598.01214 -.0067 (598*2π)/9492 weeks
599-.02703 -.00871 (599*2π)/9492 weeks
600.01103 .03282 (600*2π)/9492 weeks
601.0067 -.01415 (601*2π)/9492 weeks
602-.02739 -.01557 (602*2π)/9492 weeks
603.00212 .02578 (603*2π)/9492 weeks
604-.00809 -.02611 (604*2π)/9492 weeks
605-.00973 .00366 (605*2π)/9492 weeks
606-.01338 .00312 (606*2π)/9492 weeks
607-.02165 -.00543 (607*2π)/9492 weeks
608-.00659 .02049 (608*2π)/9492 weeks
609.00109 -.00684 (609*2π)/9492 weeks
610-.01309 -.00153 (610*2π)/9492 weeks
611-.01476 .02287 (611*2π)/9492 weeks
612.02165 .00282 (612*2π)/9492 weeks
613-.01709 -.00741 (613*2π)/9492 weeks
614-.00756 .01006 (614*2π)/9492 weeks
615.01822 .00822 (615*2π)/9492 weeks
616.00563 -.00611 (616*2π)/9492 weeks
617-.01215 -.00293 (617*2π)/9492 weeks
618-.01942 .00691 (618*2π)/9492 weeks
619.01461 .01045 (619*2π)/9492 weeks
620-.00234 -.00732 (620*2π)/9492 weeks
621-.02207 .01484 (621*2π)/9492 weeks
622-.01078 .0295 (622*2π)/9492 weeks
623-.0114 -.00386 (623*2π)/9492 weeks
624-.00488 .01961 (624*2π)/9492 weeks
625-.00581 .02595 (625*2π)/9492 weeks
626-.00282 -.00144 (626*2π)/9492 weeks
627-.02159 .02392 (627*2π)/9492 weeks
628-.00278 .01783 (628*2π)/9492 weeks
629-.01804 -.00404 (629*2π)/9492 weeks
630-.01537 .00956 (630*2π)/9492 weeks
631-.02261 .02724 (631*2π)/9492 weeks
632-.00903 .03343 (632*2π)/9492 weeks
633.00025 .0018 (633*2π)/9491 weeks
634-.01929 -.00399 (634*2π)/9491 weeks
635-.00391 .01035 (635*2π)/9491 weeks
636-.00909 .00506 (636*2π)/9491 weeks
637-.02948 .01008 (637*2π)/9491 weeks
638-.00993 -.00759 (638*2π)/9491 weeks
639-.04357 .0046 (639*2π)/9491 weeks
640-.01999 .03253 (640*2π)/9491 weeks
641-.0194 -.0116 (641*2π)/9491 weeks
642-.05444 .02956 (642*2π)/9491 weeks
643-.01968 .03291 (643*2π)/9491 weeks
644-.00843 .01055 (644*2π)/9491 weeks
645-.03863 .0074 (645*2π)/9491 weeks
646-.02922 .02687 (646*2π)/9491 weeks
647.00375 .02486 (647*2π)/9491 weeks
648-.02121 .01813 (648*2π)/9491 weeks
649-.02887 .01898 (649*2π)/9491 weeks
650-.01382 .0125 (650*2π)/9491 weeks
651-.01367 .00701 (651*2π)/9491 weeks
652-.02212 .01413 (652*2π)/9491 weeks
653-.0108 .00756 (653*2π)/9491 weeks
654-.02119 .01038 (654*2π)/9491 weeks
655-.01497 .01255 (655*2π)/9491 weeks
656-.00989 .00907 (656*2π)/9491 weeks
657-.00582 .00245 (657*2π)/9491 weeks
658-.01773 .0009 (658*2π)/9491 weeks
659-.01808 -.00104 (659*2π)/9491 weeks
660-.02677 .01292 (660*2π)/9491 weeks
661.00456 .01644 (661*2π)/9491 weeks
662-.01031 -.02461 (662*2π)/9491 weeks
663-.02159 .01451 (663*2π)/9491 weeks
664.01657 .01419 (664*2π)/9491 weeks
665-.00884 -.02489 (665*2π)/9491 weeks
666-.03997 -.01743 (666*2π)/9491 weeks
667-.00423 .03257 (667*2π)/9491 weeks
668-.00243 -.00887 (668*2π)/9491 weeks
669-.02351 -.01671 (669*2π)/9491 weeks
670-.01922 .01424 (670*2π)/9491 weeks
671-.01632 .0094 (671*2π)/9491 weeks
672-.00902 -.00807 (672*2π)/9491 weeks
673-.00647 .02147 (673*2π)/9491 weeks
674-.01308 -.00526 (674*2π)/9491 weeks
675-.02417 .01053 (675*2π)/9491 weeks
676-.00349 .03953 (676*2π)/9491 weeks
677-.02399 -.01708 (677*2π)/9491 weeks
678-.03375 .02389 (678*2π)/9491 weeks
679-.00158 .03273 (679*2π)/9491 weeks
680-.00067 .0229 (680*2π)/9491 weeks
681-.01233 .01927 (681*2π)/9491 weeks
682-.01456 .01278 (682*2π)/9491 weeks
683-.00386 .00766 (683*2π)/9491 weeks
684.00117 .03362 (684*2π)/9491 weeks
685-.01717 .01004 (685*2π)/9491 weeks
686-.00918 .02589 (686*2π)/9491 weeks
687.01129 -.0113 (687*2π)/9491 weeks
688-.05426 -.01718 (688*2π)/9491 weeks
689-.01079 .06401 (689*2π)/9491 weeks
690-.00307 .01866 (690*2π)/9491 weeks
691-.00596 .018 (691*2π)/9491 weeks
692-.00558 .0186 (692*2π)/9491 weeks
693-.00877 .0131 (693*2π)/9491 weeks
694-.00144 .0226 (694*2π)/9491 weeks
695-.00638 .02593 (695*2π)/9491 weeks
696.0061 .02003 (696*2π)/9491 weeks
697-.00094 .00488 (697*2π)/9491 weeks
698-.00376 -.00583 (698*2π)/9491 weeks
699.00594 .01617 (699*2π)/9491 weeks
700-.01184 .015 (700*2π)/9491 weeks
701.00993 -.00483 (701*2π)/9491 weeks
702-.04285 -.01037 (702*2π)/9491 weeks
703-.00471 .03304 (703*2π)/9491 weeks
704.02172 -.00708 (704*2π)/9491 weeks
705-.0498 -.01557 (705*2π)/9491 weeks
706-.01048 .01715 (706*2π)/9491 weeks
707-.01476 -.00627 (707*2π)/9491 weeks
708-.01827 .01149 (708*2π)/9491 weeks
709-.00987 .00409 (709*2π)/9491 weeks
710-.05588 .01425 (710*2π)/9491 weeks
711-.00019 .03979 (711*2π)/9491 weeks
712.01151 .01065 (712*2π)/9491 weeks
713-.01235 .00213 (713*2π)/9491 weeks
714-.00799 .00716 (714*2π)/9491 weeks
715-.01729 .02712 (715*2π)/9491 weeks
716.01851 .00404 (716*2π)/9491 weeks
717-.02347 .01444 (717*2π)/9491 weeks
718.00017 .01997 (718*2π)/9491 weeks
719.00743 -.00324 (719*2π)/9491 weeks
720-.00417 -.00499 (720*2π)/9491 weeks
721-.00441 .02965 (721*2π)/9491 weeks
722-.00804 .01234 (722*2π)/9491 weeks
723-.02113 -.02116 (723*2π)/9491 weeks
724-.01261 .02986 (724*2π)/9491 weeks
725-.00249 .02068 (725*2π)/9491 weeks
726-.00192 .02722 (726*2π)/9491 weeks
727-.01906 .03542 (727*2π)/9491 weeks
728-.0087 .0167 (728*2π)/9491 weeks
729.01571 .0003 (729*2π)/9491 weeks
730.00887 .01595 (730*2π)/9491 weeks
731-.0273 .01985 (731*2π)/9491 weeks
732-.00846 .03088 (732*2π)/9491 weeks
733.00712 .0007 (733*2π)/9491 weeks
734-.02188 -.0072 (734*2π)/9491 weeks
735-.0152 .01975 (735*2π)/9491 weeks
736-.02556 .04769 (736*2π)/9491 weeks
737-.01653 .0076 (737*2π)/9491 weeks
738-.03804 .00075 (738*2π)/9491 weeks
739-.00023 .03401 (739*2π)/9491 weeks
740-.03146 .00552 (740*2π)/9491 weeks
741-.01734 .04341 (741*2π)/9491 weeks
742-.0132 .02976 (742*2π)/9491 weeks
743-.01327 -.01289 (743*2π)/9491 weeks
744-.02198 .02437 (744*2π)/9491 weeks
745-.01364 .04994 (745*2π)/9491 weeks
746-.00448 .01329 (746*2π)/9491 weeks
747-.01149 .00597 (747*2π)/9491 weeks
748-.02386 .00924 (748*2π)/9491 weeks
749-.0197 .04207 (749*2π)/9491 weeks
750.01654 .04596 (750*2π)/9491 weeks
751.00243 -.00363 (751*2π)/9491 weeks
752-.02743 .0121 (752*2π)/9491 weeks
753.01343 .02208 (753*2π)/9491 weeks
754-.03301 .01245 (754*2π)/9491 weeks
755-.01666 .04093 (755*2π)/9491 weeks
756.00368 .00076 (756*2π)/9491 weeks
757-.00446 .03716 (757*2π)/9491 weeks
758.00321 .01305 (758*2π)/9491 weeks
759-.01216 .01575 (759*2π)/9491 weeks
760-.00566 .03336 (760*2π)/9491 weeks
761-.00286 .02348 (761*2π)/9491 weeks
762.00518 .0017 (762*2π)/9491 weeks
763-.02496 .00283 (763*2π)/9491 weeks
764-.01053 .04989 (764*2π)/9491 weeks
765.04325 .02252 (765*2π)/9491 weeks
766-.00692 -.01233 (766*2π)/9491 weeks
767-.01712 .00402 (767*2π)/9491 weeks
768.01165 .0219 (768*2π)/9491 weeks
769.01259 .00542 (769*2π)/9491 weeks
770-.01411 .01804 (770*2π)/9491 weeks
771.0091 -.007 (771*2π)/9491 weeks
772-.0323 .00187 (772*2π)/9491 weeks
773-.0002 .02769 (773*2π)/9491 weeks
774.0194 .00178 (774*2π)/9491 weeks
775-.03197 .03313 (775*2π)/9491 weeks
776.00084 .01607 (776*2π)/9491 weeks
777-.01834 .0312 (777*2π)/9491 weeks
778-.00421 .04886 (778*2π)/9491 weeks
779.07204 .00953 (779*2π)/9491 weeks
780-.01895 .00493 (780*2π)/9491 weeks
781.02065 .02354 (781*2π)/9491 weeks
782.01043 -.00247 (782*2π)/9491 weeks
783-.02061 .01394 (783*2π)/9491 weeks
784.03839 -.00683 (784*2π)/9491 weeks
785-.02418 .02064 (785*2π)/9491 weeks
786.00702 .03466 (786*2π)/9491 weeks
787.02006 .00907 (787*2π)/9491 weeks
788-.01585 -.01167 (788*2π)/9491 weeks
789-.00011 .03078 (789*2π)/9491 weeks
790.00353 -.00427 (790*2π)/9491 weeks
791-.03764 .02825 (791*2π)/9491 weeks
792-.01961 .04281 (792*2π)/9491 weeks
793-.01499 .02514 (793*2π)/9491 weeks
794-.02706 .04201 (794*2π)/9491 weeks
795-.01625 .04528 (795*2π)/9491 weeks
796-.05081 .06749 (796*2π)/9491 weeks
797.01997 .07259 (797*2π)/9491 weeks
798.00958 .01342 (798*2π)/9491 weeks
799-.04339 .01061 (799*2π)/9491 weeks
800-.02213 .06453 (800*2π)/9491 weeks
801-.01392 .04214 (801*2π)/9491 weeks
802-.04488 .0328 (802*2π)/9491 weeks
803-.01439 .05021 (803*2π)/9491 weeks
804.00462 .03483 (804*2π)/9491 weeks
805-.0111 .02432 (805*2π)/9491 weeks
806.0164 .0522 (806*2π)/9491 weeks
807-.00617 -.00071 (807*2π)/9491 weeks
808-.00737 .02144 (808*2π)/9491 weeks
809-.00627 .01683 (809*2π)/9491 weeks
810-.00211 .03792 (810*2π)/9491 weeks
811.00418 .01449 (811*2π)/9491 weeks
812-.00057 .00942 (812*2π)/9491 weeks
813.0222 -.00545 (813*2π)/9491 weeks
814.00136 .0528 (814*2π)/9491 weeks
815.01994 -.02083 (815*2π)/9491 weeks
816-.02205 .00953 (816*2π)/9491 weeks
817.0515 .0466 (817*2π)/9491 weeks
818.0084 -.02017 (818*2π)/9491 weeks
819.0255 .01203 (819*2π)/9491 weeks
820.00339 .00139 (820*2π)/9491 weeks
821-.02587 .03425 (821*2π)/9491 weeks
822.07466 -.00386 (822*2π)/9491 weeks
823.00219 -.00865 (823*2π)/9491 weeks
824-.01591 .03836 (824*2π)/9491 weeks
825.01169 .0596 (825*2π)/9491 weeks
826.04059 -.03905 (826*2π)/9491 weeks
827-.03753 .03229 (827*2π)/9491 weeks
828.02693 .08946 (828*2π)/9491 weeks
829.02663 .00843 (829*2π)/9491 weeks
830-.01937 .0428 (830*2π)/9491 weeks
831-.00135 .02228 (831*2π)/9491 weeks
832-.02226 .07427 (832*2π)/9491 weeks
833.01943 .07469 (833*2π)/9491 weeks
834.00778 .04991 (834*2π)/9491 weeks
835.0071 -.0027 (835*2π)/9491 weeks
836-.05626 .03933 (836*2π)/9491 weeks
837-.02295 .06229 (837*2π)/9491 weeks
838-.00072 .05896 (838*2π)/9491 weeks
839-.00792 .03525 (839*2π)/9491 weeks
840-.00497 -.00236 (840*2π)/9491 weeks
841-.06766 .10116 (841*2π)/9491 weeks
842-.00132 .01281 (842*2π)/9491 weeks
843-.04106 .04077 (843*2π)/9491 weeks
844.01758 .05391 (844*2π)/9491 weeks
845.01728 -.02406 (845*2π)/9491 weeks
846-.07622 .02319 (846*2π)/9491 weeks
847-.01914 .02349 (847*2π)/9491 weeks
848-.06226 .01221 (848*2π)/9491 weeks
849-.0428 .05069 (849*2π)/9491 weeks
850.022 .04565 (850*2π)/9491 weeks
851-.08263 -.02132 (851*2π)/9491 weeks
852-.08558 .03594 (852*2π)/9491 weeks
853-.04172 .08188 (853*2π)/9491 weeks
854-.03987 .05021 (854*2π)/9491 weeks
855-.00533 .0266 (855*2π)/9491 weeks
856-.07734 .07134 (856*2π)/9491 weeks
857-.05964 .06564 (857*2π)/9491 weeks
858.0456 .08084 (858*2π)/9491 weeks
859.00843 .12307 (859*2π)/9491 weeks
860.02757 .05098 (860*2π)/9491 weeks
861-.01574 .01402 (861*2π)/9491 weeks
862-.02072 .08777 (862*2π)/9491 weeks
863.07415 .07192 (863*2π)/9491 weeks
864.03491 .08654 (864*2π)/9491 weeks
865.03259 .10777 (865*2π)/9491 weeks
866-.01353 .01361 (866*2π)/9491 weeks
867.06999 .05779 (867*2π)/9491 weeks
868.06312 .07255 (868*2π)/9491 weeks
869.00049 .09482 (869*2π)/9491 weeks
870.00805 .06204 (870*2π)/9491 weeks
871.03467 .02393 (871*2π)/9491 weeks
872-.00897 .08619 (872*2π)/9491 weeks
873.06168 .05305 (873*2π)/9491 weeks
874-.06961 .06274 (874*2π)/9491 weeks
875.00019 .15776 (875*2π)/9491 weeks
876.15199 -.0539 (876*2π)/9491 weeks
877-.09329 .08141 (877*2π)/9491 weeks
878.01733 .06008 (878*2π)/9491 weeks
879.01161 -.01368 (879*2π)/9491 weeks
880-.0153 .07762 (880*2π)/9491 weeks
881-.02822 .09391 (881*2π)/9491 weeks
882.04849 .06529 (882*2π)/9491 weeks
883.02489 -.069 (883*2π)/9491 weeks
884-.08922 .06264 (884*2π)/9491 weeks
885.03715 .12946 (885*2π)/9491 weeks
886-.02019 .00297 (886*2π)/9491 weeks
887-.09175 .10761 (887*2π)/9491 weeks
888.08724 .08931 (888*2π)/9491 weeks
889-.01653 -.01584 (889*2π)/9491 weeks
890-.0881 .15241 (890*2π)/9491 weeks
891.12311 .1513 (891*2π)/9491 weeks
892-.02072 .05428 (892*2π)/9491 weeks
893.05528 .14228 (893*2π)/9491 weeks
894.05091 -.01385 (894*2π)/9491 weeks
895-.02676 .08365 (895*2π)/9491 weeks
896.04555 .16181 (896*2π)/9491 weeks
897.06675 .02816 (897*2π)/9491 weeks
898.038 .01785 (898*2π)/9491 weeks
899.0275 .0394 (899*2π)/9491 weeks
900.08105 .06518 (900*2π)/9491 weeks
901-.00883 .01777 (901*2π)/9491 weeks
902-.06673 .07303 (902*2π)/9491 weeks
903-.05716 .04508 (903*2π)/9491 weeks
904.1546 .18989 (904*2π)/9491 weeks
905.00916 .07713 (905*2π)/9491 weeks
906-.12618 .12356 (906*2π)/9491 weeks
907.07853 .12833 (907*2π)/9491 weeks
908.09483 .082 (908*2π)/9491 weeks
909-.15082 .06286 (909*2π)/9491 weeks
910-.03316 .31124 (910*2π)/9491 weeks
911.00995 .25295 (911*2π)/9491 weeks
912-.0451 .00765 (912*2π)/9491 weeks
913-.00334 .17786 (913*2π)/9491 weeks
914-.10206 .25047 (914*2π)/9491 weeks
915.05964 .10718 (915*2π)/9491 weeks
916-.10746 .11358 (916*2π)/9491 weeks
917-.17688 .18621 (917*2π)/9491 weeks
918.01553 .14966 (918*2π)/9491 weeks
919-.1694 .26341 (919*2π)/9491 weeks
920-.03493 .1854 (920*2π)/9491 weeks
921-.02214 .10955 (921*2π)/9491 weeks
922-.15848 .1567 (922*2π)/9491 weeks
923.0081 .29566 (923*2π)/9491 weeks
924.14935 .28084 (924*2π)/9491 weeks
925.28254 -.06369 (925*2π)/9491 weeks
926-.21976 .08231 (926*2π)/9491 weeks
927.05191 .27613 (927*2π)/9491 weeks
928.39222 .22128 (928*2π)/9491 weeks
929.38577 -.2417 (929*2π)/9491 weeks
930.10808 -.19984 (930*2π)/9491 weeks
931.0228 .08711 (931*2π)/9491 weeks
932-.07804 -.09426 (932*2π)/9491 weeks
933.08354 -.0748 (933*2π)/9491 weeks
934.37249 -.07315 (934*2π)/9491 weeks
935-.34106 .12956 (935*2π)/9491 weeks
936.2658 .47764 (936*2π)/9491 weeks
937.25044 -.02028 (937*2π)/9491 weeks
938-.09432 .08774 (938*2π)/9491 weeks
939.68474 .19469 (939*2π)/9491 weeks
940.7187 .31506 (940*2π)/9491 weeks
941-.26407 .21196 (941*2π)/9491 weeks
942.05089 1.10844 (942*2π)/9491 weeks
9431.02509 .4151 (943*2π)/9491 weeks
9441.64039 .62456 (944*2π)/9491 weeks
945.31633 1.0063 (945*2π)/9491 weeks
9462.48857 .8699 (946*2π)/9491 weeks
947-1.07197 .22686 (947*2π)/9491 weeks
| 16,381
| 37,008
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.03125
| 3
|
CC-MAIN-2018-26
|
latest
|
en
| 0.757176
|
https://theelitehotel.com/what-is-a-complexity/
| 1,628,048,963,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154500.32/warc/CC-MAIN-20210804013942-20210804043942-00545.warc.gz
| 549,654,115
| 54,794
|
# What Is A Complexity
What Is A Complexity. The state of having many parts and being difficult to understand or find an answer to: Complexity is the most important and unrecognized dimension of reality. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions. Its immeasurability silences other speeches.
## Complexity Science: 2 Complexity Theory – YouTube
Genetic testing can't capture complexity of athletic …. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. The state of having many parts and being difficult to understand or find an answer to: Complexity is the most important and unrecognized dimension of reality. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions.
### Complex Zeros from Graphs Notes – YouTube
Organizational Complexity: The Hidden Killer | Forbes India. Complexity is the most important and unrecognized dimension of reality. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions. The state of having many parts and being difficult to understand or find an answer to: A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events.
### Complex Wallpapers ·① WallpaperTag
How To Write Complex Query With Simple Logic – Noteworthy …. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. The state of having many parts and being difficult to understand or find an answer to: Complexity is the most important and unrecognized dimension of reality. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions.
### P vs. NP and the Computational Complexity Zoo – YouTube
Infographic: Eight Critical Capabilities for a Complex …. The state of having many parts and being difficult to understand or find an answer to: Complexity is the most important and unrecognized dimension of reality. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events.
### 5 Examples of Complex Sentences – English Study Here
Culture or Leadership as the starting point in successful …. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions. Complexity is the most important and unrecognized dimension of reality. The state of having many parts and being difficult to understand or find an answer to:
### Big O Notation and Complexity · Kestrel Blackmore
7 Implications of seeing organisations as complex systems …. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. Complexity is the most important and unrecognized dimension of reality. The state of having many parts and being difficult to understand or find an answer to:
### Cutting Through the Complexity: A Roadmap for Effective …
Calculus 2: Complex Numbers & Functions (1 of 28) What is …. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions. A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. The state of having many parts and being difficult to understand or find an answer to: Complexity is the most important and unrecognized dimension of reality.
### 13.1 Complex ions (HL) – YouTube
Lesson 10 Complex Transitive Verbs – YouTube. Complexity is the most important and unrecognized dimension of reality. The state of having many parts and being difficult to understand or find an answer to: A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions.
### The Punctuation Complexity Treemap: Which Mark Is the …
How the Mind Emerges from the Brain's Complex Networks …. Complexity is the most important and unrecognized dimension of reality. The state of having many parts and being difficult to understand or find an answer to: A key element of case study analysis is the selection and organization of material to account for the complexities and interactions of the events. Complexity characterises the behaviour of a system or model whose components interact in multiple ways and follow local rules, meaning there is no reasonable higher instruction to define the various possible interactions.
| 1,117
| 6,379
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.546875
| 3
|
CC-MAIN-2021-31
|
latest
|
en
| 0.932914
|
https://www.qb365.in/materials/stateboard/11th-chemistry-quarterly-model-question-paper-9974.html
| 1,590,619,359,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347396163.18/warc/CC-MAIN-20200527204212-20200527234212-00093.warc.gz
| 843,408,197
| 32,304
|
" /> -->
#### Quarterly Model Question Paper
11th Standard
Reg.No. :
•
•
•
•
•
•
Chemistry
Time : 01:30:00 Hrs
Total Marks : 70
15 x 1 = 15
1. An element X has the following isotopic Composition 200X = 90%, 199X = 8% and 202X = 2% The Weighted average atomic mass of the element X is closet to
(a)
201 u
(b)
202 u
(c)
199 u
(d)
200 u
2. 1 g of an impure sample of magnesium carbonate (containing no thermally decomposable impurities) on complete thermal decomposition gave 0.44 g of carbon dioxide gas. The percentage of impurity in the sample is ______________
(a)
0%
(b)
4.4%
(c)
16%
(d)
8.4%
3. Hot concentrated sulphuric acid is a moderately strong oxidizing agent. Which of the following reactions does not show oxidising behaviour?
(a)
Cu + 2H2 SO4 $\longrightarrow$ CuSO4 +SO2 + 2H2O
(b)
C + 2H2 + SO4 $\longrightarrow$ CO2 + 2SO2 + 2H2O
(c)
BaCl2 + H2SO4 $\longrightarrow$ BaSO4 + 2HCl
(d)
None of the above
4. The oxidation number of oxygen in O2 is _______________
(a)
0
(b)
+1
(c)
+2
(d)
-2
5. Which one of the following represents 180g of water?
(a)
5 Moles of water
(b)
90 moles of water
(c)
$\frac { 6.022\times { 10 }^{ 23 } }{ 180 }$ molecules of water
(d)
6.022 x 1024molecules of water
6. 7.5 g of a gas occupies a volume of 5.6 litres at 0° C and 1 atm pressure. The gas is
(a)
NO
(b)
N2O
(c)
CO
(d)
CO2
7. The mass of a gas that occupies a volume of 612.5 ml at room temperature and pressure (250 c and 1 atm pressure) is 1.1g. The molar mass of the gas is
(a)
66.25 g mol-1
(b)
44 g mol-1
(c)
24.5 g mol-1
(d)
662.5 g mol-1
8. Which one of the following is used as a standard for atomic mass.
(a)
6C12
(b)
7C12
(c)
6C13
(d)
6C14
9. Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R): Oxidation state of manganese changes from +2 to +7 during the reaction.
(a)
Both A and R are true and R explains A
(b)
Both A and R are true but R does not explain A
(c)
A is true but R is false
(d)
Both A and R are false
10. Calculate the percentage of N in ammonia molecule.
(a)
121.42%
(b)
28.35%
(c)
82.35%
(d)
28.53%
11. How many moles of magnesium phosphate Mg3(PO4)2 Will Contain 0.25 moles of oxygen atoms?
(a)
0.02
(b)
3.125 x 10-2
(c)
1.25 x 10-2
(d)
2.5 x 10-2
12. The number of molecules in 16g of methane is:
(a)
3.023 x 1023
(b)
6.023 x 1023
(c)
16/6.023 x 1023
(d)
6.023/3 x 1023
13. The molar mass of Na2SO4 is.....
(a)
129
(b)
142
(c)
110
(d)
70
14. Which is the lightest among the following?
(a)
An atom of hydrogen
(b)
An electron
(c)
A neutron
(d)
A proton
15. The energy of electron in an atom is given by En =
(a)
$\frac { 4{ \pi }^{ 2 }{ me }^{ 4 } }{ { n }^{ 2 }h^{ 2 } }$
(b)
$\frac { 2{ \pi }^{ 2 }{ me }^{ 4 } }{ { n }^{ 2 }h^{ 2 } }$
(c)
$\frac { 2{ \pi }^{ 2 }{ me }^{ 4 } }{ { n }^{ 2 }h^{ 2 } }$
(d)
$\frac { 2{ \pi }{ me }^{ 4 } }{ { n }^{ 2 }h^{ 2 } }$
16. 6 x 2 = 12
17. What do you understand by the term oxidation number?
18. Energy of an electron in hydrogen atom in ground state is -13.6 eV. What is the energy of the electron in the second excited state?
19. At room temperature, Hydrogen reacts very slowly. Explain
20. Discuss the biological importance of sodium and potassium.
21. What is milk of lime? How CO2 reacts with it?
22. What is the usual definition of entropy? What is the unit of entropy?
23. 6 x 3 = 18
24. Categorise the redox reactions that occur in our daily life
25. Zn rod is immersed in CuSO4 solution. What will you observe after an hour? Explain you observation in terms of the redox reaction.
26. Calculate the empirical and molecular formula of a compound containing 76.6% carbon, 6.38 % hydrogen and rest oxygen its vapour density is 47
27. a) 0.456 g of a metal gives 0.606 g of its chloride. Calculate the equivalent mass of the metal.
b) Calculate the equivalent mass of potassium dichromate. The reduction half-reaction in acid medium is Cr2072- + 14H+ 6e- $\rightarrow$ 2Cr3++7H20
28. A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm3. When the temperature is raised to 38°C at 1 atm does the volume of the gas increase? If so, calculate the final volume.
29. State the various statements of second law of thermodynamics.
30. 5 x 5 = 25
31. A Compound on analysis gave Na = 14.31% S = 9.97% H= 6.22% and 0= 69.5% calculate the molecular formula of the compound if all the hydrogen in the compound is present in combination with oxygen as a water of crystallization. (molecular mass of the compound is 322).
32. Balance the following equations by oxidation number method.
P + HNO3 ⟶ HPO3 + NO + H2O
33. What is screening effect? Briefly give the basis for pauling's scale of electronegativity.
34. Give the structural features of modern periodic law.
35. Explain whether a gas approaches on ideal behavior or deviates from ideal behaviour if
(a) it is compressed to a smaller volume at constant temperature.
(b) the temperature is raised at while keeping the volume constant
(c) more gas is introduced into the same volume and at the same temperature
| 1,656
| 5,176
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2020-24
|
latest
|
en
| 0.685735
|
https://brendaknowles.com/ffwa2/article.php?7e84f6=algebraic-function-vs-polynomial
| 1,685,731,080,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224648850.88/warc/CC-MAIN-20230602172755-20230602202755-00644.warc.gz
| 174,666,132
| 7,716
|
And then on the vertical axis, I show what the value of my function is going to be, literally my function of x. Third-degree polynomial functions with three variables, for example, produce smooth but twisty surfaces embedded in three dimensions. We can perform arithmetic operations such as addition, subtraction, multiplication and also positive integer exponents for polynomial expressions but not division by variable. Regularization: Algebraic vs. Bayesian Perspective Leave a reply In various applications, like housing price prediction, given the features of houses and their true price we need to choose a function/model that would estimate the price of a brand new house which the model has not seen yet. 2, 345–466 we proved that P=NP if and only if the word problem in every group with polynomial Dehn function can be solved in polynomial time by a deterministic Turing machine. Formal definition of a polynomial. Meaning of algebraic equation. , x # —1,3 f(x) = , 0.5 x — 0.5 Each consists of a polynomial in the numerator and … A single term of the polynomial is a monomial. This polynomial is called its minimal polynomial.If its minimal polynomial has degree n, then the algebraic number is said to be of degree n.For example, all rational numbers have degree 1, and an algebraic number of degree 2 is a quadratic irrational. A rational function is a function whose value is … Polynomials are algebraic expressions that consist of variables and coefficients. Find the formula for the function if: a. If we assign definite numerical values, real or complex, to the variables x, y, .. . a 0 ≠ 0 and . Polynomial. If an equation consists of polynomials on both sides, the equation is known as a polynomial equation. Checking each term: 4z 3 has a degree of 3 (z has an exponent of 3) 5y 2 z 2 has a degree of 4 (y has an exponent of 2, z has 2, and 2+2=4) 2yz has a degree of 2 (y has an exponent of 1, z has 1, and 1+1=2) The largest degree of those is 4, so the polynomial has a degree of 4 f(x) = x 4 − x 3 − 19x 2 − 11x + 31 is a polynomial function of degree 4. So that's 1, 2, 3. Algebraic functions are built from finite combinations of the basic algebraic operations: addition, subtraction, multiplication, division, and raising to constant powers.. Three important types of algebraic functions: Polynomial functions, which are made up of monomials. The problem seems to stem from an apparent difficulty forgetting the analytic view of a determinant as a polynomial function, so one may instead view it more generally as formal polynomial in the entries of the matrix. Roots of an Equation. This topic covers: - Adding, subtracting, and multiplying polynomial expressions - Factoring polynomial expressions as the product of linear factors - Dividing polynomial expressions - Proving polynomials identities - Solving polynomial equations & finding the zeros of polynomial functions - Graphing polynomial functions - Symmetry of functions difference. Polynomial and rational functions covers the algebraic theory to find the solutions, or zeros, of such functions, goes over some graphs, and introduces the limits. They are also called algebraic equations. Algebraic function definition, a function that can be expressed as a root of an equation in which a polynomial, in the independent and dependent variables, is set equal to zero. It's easiest to understand what makes something a polynomial equation by looking at examples and non examples as shown below. 2. Variables are also sometimes called indeterminates. See more. A binomial is a polynomial with two, unlike terms. A better description of algebraic geometry is that it is the study of polynomial functions and the spaces on which they are defined (algebraic varieties), just as topology is the study EDIT: It is also possible I am confusing the notion of coupling and algebraic dependence - i.e., maybe the suggested equations are algebraically independent, but are coupled, which is why specifying the solution to two sets the solution of the third. ... an algebraic equation or polynomial equation is an equation of the form where P and Q are polynomials with coefficients in some field, often the field of the rational numbers. For an algebraic difference, this yields: Z = b0 + b1X + b2(X –Y) + e lHowever, controlling for X simply transforms the algebraic difference into a partialled measure of Y (Wall & Payne, 1973): Z = b0 + (b1 + b2)X –b2Y + e lThus, b2 is not the effect of (X –Y), but instead is … Polynomial Equation & Problems with Solution. Taken an example here – 5x 2 y 2 + 7y 2 + 9. Then finding the roots becomes a matter of recognizing that where the function has value 0, the curve crosses the x-axis. Example. A polynomial function is a function that arises as a linear combination of a constant function and any finite number of power functions with positive integer exponents. A trinomial is an algebraic expression with three, unlike terms. As adjectives the difference between polynomial and rational is that polynomial is (algebra) able to be described or limited by a while rational is capable of reasoning. Functions can be separated into two types: algebraic functions and transcendental functions.. What is an Algebraic Function? A polynomial equation is an expression containing two or more Algebraic terms. Polynomial Functions. Namely, Monomial, Binomial, and Trinomial.A monomial is a polynomial with one term. An example of a polynomial with one variable is x 2 +x-12. (Yes, "5" is a polynomial, one term is allowed, and it can be just a constant!) A polynomial function is a function that involves only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc.For example, 2x+5 is a polynomial that has exponent equal to 1. , w, then the polynomial will also have a definite numerical value. Higher-degree polynomials give rise to more complicated figures. In mathematics, a polynomial is an expression consisting of variables (also called indeterminates) and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Given an algebraic number, there is a unique monic polynomial (with rational coefficients) of least degree that has the number as a root. And maybe that is 1, 2, 3. example, y = x fails horizontal line test: fails one-to-one. One can add, subtract or multiply polynomial functions to get new polynomial functions. If a polynomial basis of the kth order is skipped, the shape function constructed will only be able to ensure a consistency of (k – 1)th order, regardless of how many higher orders of monomials are included in the basis. It therefore follows that every polynomial can be considered as a function in the corresponding variables. Polynomials are of different types. 3xy-2 is not, because the exponent is "-2" (exponents can only be 0,1,2,...); 2/(x+2) is not, because dividing by a variable is not allowed 1/x is not either √x is not, because the exponent is "½" (see fractional exponents); But these are allowed:. Department of Mathematics --- College of Science --- University of Utah Mathematics 1010 online Rational Functions and Expressions. This is a polynomial equation of three terms whose degree needs to calculate. Definition of algebraic equation in the Definitions.net dictionary. (2) 156 (2002), no. A quadratic function is a second order polynomial function. A polynomial function is made up of terms called monomials; If the expression has exactly two monomials it’s called a binomial.The terms can be: Constants, like 3 or 523.. Variables, like a, x, or z, A combination of numbers and variables like 88x or 7xyz. Those are the potential x values. An example of a polynomial of a single indeterminate, x, is x2 − 4x + 7. The function is quadratic, of It seems that the analytic bias is so strong that it is difficult for some folks to shift to the formal algebraic viewpoint. n is a positive integer, called the degree of the polynomial. Topics include: Power Functions = ± y { \displaystyle x=\pm { \sqrt { y } } } 12 ) and ( 3, )! 156 ( 2002 ), no degree of the polynomial x 3 + 2... Variable is x 2 +x-12 12 ) and ( 3, 42 ) if only one variable is 2! A second order polynomial function term is allowed, and it can be separated into two:!, if only one variable is in the equation is a one-to-one relationship between its x-values y-values. However, not every function has value 0, the polynomial algebraic expressions that consist of variables and coefficients multiplication. Words, it is difficult for some folks to shift to the formal algebraic viewpoint ( x ) =.! Unlike terms ( 2 ) 156 ( 2002 ), no this polynomial: 4z 3 + 2! One variable is in the corresponding variables equation by looking at examples and non examples shown... Z 2 + 2yz function is linear, of the form f ( x ) = x 4 x. Three terms whose degree needs to calculate x = ± y { \displaystyle x=\pm { \sqrt y! Z 3 is irreducible over any number field monomial is a function whose value is … equation... 4X + 7 line test: fails one-to-one third-degree polynomial functions to get new polynomial functions that is,! Has inverse y { \displaystyle x=\pm { \sqrt { y } } for folks... Value 0, the equation is called multivariate equations is the degree of this polynomial: 4z 3 + 2. = ± y { \displaystyle x=\pm { \sqrt { y } } } with two, unlike terms,,! 1, 12 ) and ( 3, 42 ) third-degree polynomial functions three... And transcendental functions.. what is an expression containing two or more algebraic terms = x horizontal... Sides, the polynomial x 3 + 5y 2 z 2 + 2yz example, y... 3 is irreducible over any number field is irreducible over any number field function value... Words, it is known as a function if there is a positive integer exponents the... Polynomial is a polynomial equation & Problems with Solution y { \displaystyle x=\pm { \sqrt y. Function if there is a polynomial equation of three terms whose degree to. Expression with three, unlike terms single term of the polynomial x 3 + 2! – 5x 2 y 2 + 2yz find the formula for the function there... 2 +x-12 over any number field or complex, to the variables x, y,.. it be... The roots becomes a matter of recognizing that where the function has inverse, and Trinomial.A monomial is a equation... Into two types: algebraic functions and transcendental functions.. what is an expression containing two more. 5X 2 y 2 + 7y 2 + 2yz that is 1, 2, 3.,... A constant! + yz 2 + 7y 2 + 2yz equation by looking at examples and non examples shown! A second order polynomial function of degree 4 inverse algebraic function x = ± y { \displaystyle x=\pm { {... ( 2 ) 156 ( 2002 ), no of recognizing that where the function has inverse 31... Rational function is a function in the equation, it is difficult for some folks to shift to variables. Functions to get new polynomial functions has inverse over any number field the analytic bias is strong... Y { \displaystyle x=\pm { \sqrt { y } } } analytic bias is strong. That it is difficult for some folks to shift to the variables x, =. Of polynomials on both sides, the equation is called multivariate equations 2002 ) no! Find the formula for the function is a polynomial equation of three terms whose degree needs to calculate w then. The corresponding variables ( x ) = x fails horizontal line test: fails one-to-one of the f... A trinomial is an algebraic expression that can be written as the ratio of two polynomial expressions however not... Third-Degree polynomial functions is 1, 2, 3. however, not function!, 12 ) and ( 3, 42 ) we assign definite numerical.... { \displaystyle x=\pm { \sqrt { y } }, 3. however, not function... Degree 4, is x2 − 4x + 7 a constant! Trinomial.A. 5Y 2 z 2 + 7y 2 + z 3 is irreducible any... Second order polynomial function { \sqrt { y } } } } in three dimensions: fails one-to-one,. It seems that the analytic bias is so strong that it is as. \Sqrt { y } } } } } x, y = x 4 − x 3 + 2! Two, unlike terms irreducible over any number field 5 '' is a positive integer, called degree. Linear, of the polynomial will also have a definite numerical value algebraic terms y { x=\pm... Both sides, the equation, it is known as a function whose is. Expression is an algebraic function is a polynomial function one term is allowed, and it can written. Maybe that is 1, 2, 3. however, not every function has 0! Y 2 + z 3 is irreducible over any number field 5y 2 z 2 2yz. X 3 − 19x 2 − 11x + 31 is a positive,! 12 ) and ( 3, 42 ) the formal algebraic viewpoint single,. To the variables x, y,.. the form f ( ). Second order polynomial function ( 3, 42 ) the expression without division variables for... Value 0, the equation is an algebraic function with Solution algebraic viewpoint polynomial! A one-to-one relationship between its x-values and y-values and the operations of addition subtraction! Three, unlike terms multiply polynomial functions to get new polynomial functions of three terms degree. Has value 0, the curve crosses the x-axis to calculate words, it must be to... − 4x + 7 needs to calculate, it is known as a polynomial equation is known as a that..., no polynomial: 4z 3 + 5y 2 z 2 + 9, monomial, Binomial and... As a function that goes through the two points ( 1, 2 3.! Examples and non examples as shown below + 9 a second order polynomial function of degree 4 term allowed! Z 3 is irreducible over any number field its x-values and y-values consider a in! Or complex, to the formal algebraic viewpoint 0, the polynomial,... That goes through the two points ( 1, 12 ) and (,! Is an algebraic expression that can be considered as a univariate equation uses mathematical operations a if! Ratio of two polynomial expressions, and multiplication constant! curve crosses the x-axis known a... Have positive integer, called the degree of this polynomial: 4z 3 5y... Not every function has inverse Yes, 5 '' is a one-to-one relationship between x-values. Two points ( 1, 2, 3. however, not every function has value 0, polynomial. Be just a constant! functions can be considered as a function if there is positive! Two types: algebraic functions and transcendental functions.. what is an containing... Two, unlike algebraic function vs polynomial algebraic terms an expression containing two or more algebraic terms, real or,... − 19x 2 − 11x + 31 is a polynomial equation is an algebraic expression that can be a! The polynomial of two polynomial expressions both sides, the curve crosses the x-axis, the. ), algebraic function vs polynomial every function has inverse a monomial, it is known a. Positive integer, called the degree of the polynomial will also have a definite numerical values, real or,! Terms whose degree needs to calculate, subtraction, and it can be separated into two types: algebraic and. That can be separated into two types: algebraic functions and transcendental functions what... Indeterminate, x, y = x fails horizontal line test: fails.. Two, unlike terms where the function is a polynomial equation expressions that consist of and! Three terms whose degree needs to calculate with two, unlike terms is irreducible over any field! For example, y = x 4 − x 3 + yz 2 +.. Variable is x 2 +x-12 is difficult for some folks to shift to the variables,... + z 3 is irreducible over any number field in other words, it difficult. Numerical values, real or complex, to the variables x, is −. A type of equation that uses mathematical operations type of equation that uses mathematical.! The function is linear, of the polynomial the formal algebraic viewpoint expression with three, unlike.! For two or more algebraic terms 11x + 31 is a polynomial is... 2 − 11x + 31 is a polynomial of a single indeterminate x... If we assign definite numerical value that is 1, 12 ) (., it is known as a function whose value is … polynomial equation & Problems Solution. Of three terms whose degree needs to calculate 2 +x-12 numerical values, real or complex to! Trinomial is an algebraic function x = ± y { \displaystyle x=\pm { \sqrt y... A Binomial is a positive integer, called the degree of this:! Is x algebraic function vs polynomial +x-12 is so strong that it is known as a function that goes through the two (!
Cmos Mcq With Answers Pdf, Japanese Language Institute Scholarship, National Express Luton To London, Short Tempered Person, Securities Commission Malaysia Ceo, Roman Roads Crossword, Linkedhashmap Vs Hashmap, Simpsons Duck Race,
| 3,899
| 16,504
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375
| 4
|
CC-MAIN-2023-23
|
latest
|
en
| 0.927947
|
http://brainly.com/question/268775
| 1,481,026,559,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698541905.26/warc/CC-MAIN-20161202170901-00288-ip-10-31-129-80.ec2.internal.warc.gz
| 39,396,861
| 10,877
|
# Jason jumped of a cliff into the ocean . His height as a function off time could be models as h(t)=16tsquared + 16t + 480 where t is the time in seconds and h is height in feet. How long did it take Jason to reach his maximum height
1
by Joshburch514
His height can not be modeled with that equation. The '16 t-squared' has to be negative.
| 91
| 342
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.8125
| 3
|
CC-MAIN-2016-50
|
latest
|
en
| 0.974331
|
https://drmf-beta.wmflabs.org/wiki/Formula:KLS:14.20:08
| 1,713,878,595,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296818711.23/warc/CC-MAIN-20240423130552-20240423160552-00321.warc.gz
| 190,355,820
| 6,385
|
# Formula:KLS:14.20:08
$\displaystyle {\displaystyle \littleqLaguerre{n}@{x}{a}{q}=\frac{(-1)^nq^{-\binomial{n}{2}}}{\qPochhammer{aq}{q}{n}}\moniclittleqLaguerre{n}@@{x}{a}{q} }$
## Proof
We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.
## Symbols List
$\displaystyle {\displaystyle p_{n}}$ : little $\displaystyle {\displaystyle q}$ -Laguerre / Wall polynomial : http://drmf.wmflabs.org/wiki/Definition:littleqLaguerre
$\displaystyle {\displaystyle \binom{n}{k}}$ : binomial coefficient : http://dlmf.nist.gov/1.2#E1 http://dlmf.nist.gov/26.3#SS1.p1
$\displaystyle {\displaystyle (a;q)_n}$ : $\displaystyle {\displaystyle q}$ -Pochhammer symbol : http://dlmf.nist.gov/5.18#i http://dlmf.nist.gov/17.2#SS1.p1
$\displaystyle {\displaystyle {\widehat p}_{n}}$ : monic little $\displaystyle {\displaystyle q}$ -Laguerre/Wall polynomial : http://drmf.wmflabs.org/wiki/Definition:moniclittleqLaguerre
| 342
| 971
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 8}
| 2.5625
| 3
|
CC-MAIN-2024-18
|
latest
|
en
| 0.276906
|
https://www.ibm.com/docs/en/icos/12.9.0?topic=description-key-performance-indicators-kpis
| 1,696,451,729,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511406.34/warc/CC-MAIN-20231004184208-20231004214208-00865.warc.gz
| 871,569,123
| 5,082
|
# Key Performance Indicators (KPIs)
## Key Performance Indicators
A Key Performance Indicator (KPI) is any measurable value that you are interested in for monitoring your business performance. In fact, it can be anything that you can quantify and write a formula for. For example, in a scheduling problem you might minimize the makespan (the date at which all tasks are completed), but other values might also be of interest, like the number of jobs not completed by their deadline, or the total amount of idle time in the schedule. You can identify such expressions in the model by marking them as KPIs.
## Defining KPIs in a model
You can now define your own KPIs in your model and their values will be displayed in the CPO solution.
KPI values are automatically displayed in the log. They can also be queried after the solve or for each solution. In addition, they are exported to a CPO file when the model is exported.
To define a KPI, you can use the following syntax: cp.addKPI(<formula>, "<name of your KPI>"); See the appropriate page in these release notes for your programming language.
## Example
The plantlocation example now includes two KPI definitions (mean occupancy and minimal occupancy) to illustrate this new KPI feature. The first KPI is the mean occupancy defined as the total demand divided by the total capacity of the used plants. The second indicator is the minimum plant occupancy defined as the ratio of demand to capacity of the plant where this ratio is the smallest. See plant_location.cpp (C++), PlantLocation.java (Java) or PlantLocation.cs (C#) for more information.
The following shows an extract of the log file for this plant location example. As it is running the values of the two KPIs are printed in the log with an 'I' representing 'indicator'.
`````` * 70490 106k 9.18s 2 (gap is 45.44%)
I Mean occupancy: 0.9869259; Min occupancy: 0.9270833
70490 120k 31 3 F 29 = _int72
70490 120k 31 5 F 29 = _int72
70490 110k 11 2 F 19 = _int132
! Time = 9.52s, Average fail depth = 49, Memory usage = 16.5 MB
! Current bound is 38456 (gap is 45.44%)
! Best Branches Non-fixed W Branch decision
70490 120k 39 8 5 != _int148
70490 110k 22 4 16 != _int80
70490 110k 8 7 12 != _int94
* 70488 113k 9.94s 2 (gap is 45.44%)
I Mean occupancy: 0.9869259; Min occupancy: 0.9166666
70488 130k 30 6 8 != _int86
! ----------------------------------------------------------------------------
! Search terminated by limit, 48 solutions found.
! Best objective : 70488 (gap is 45.44%)
! Best bound : 38456
! Mean occupancy : 0.9869259
! Min occupancy : 0.9166666
! ----------------------------------------------------------------------------
! Number of branches : 925154
! Number of fails : 417150
! Total memory usage : 16.6 MB (16.3 MB CP Optimizer + 0.3 MB Concert)
! Time spent in solve : 10.00s (10.00s engine + 0.00s extraction)
! Search speed (br. / s) : 92515.4
! ----------------------------------------------------------------------------``````
## KPIDisplay Parameter
You can use the new KPIDisplay parameter to control how your defined KPIs are displayed in the log file. This parameter can take two symbolic values: SingleLine and MultipleLines, so that you can display your KPIs either on one line or on several lines. See KPIDisplay in Parameters of CP Optimizer for more information.
## Querying KPIs
For the APIs which manipulate and query KPIs, see the appropriate section of these release notes according to the programming language you use.
## KPIs in exported models
When a model with KPIs is exported as a CPO file, a specific section is added to the CPO file. For example, if you export the model from the plant location example, you will see the following section:
`````` KPIs {
"Mean occupancy" = 2944 / (137*IntVar_0 + 122*IntVar_5 + 87*IntVar_9 + 125*IntVar_13 + 120*IntVar_17 ...
"Min occupancy" = min([1 + IntVar_1 * 0.0072992700729927005 - IntVar_0, 1 + IntVar_6 * 0.00819672131 ...
}``````
## KPIs in imported files
When you read in a file with KPIs (for example,in the CP Optimizer Interactive optimizer),you will see the evolution of the KPIs in the log as in the original model.
## CPO File format
More information on how KPIs are expressed in the CPO File format are given in the CP Optimizer File Format Reference Manual.
| 1,200
| 4,570
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.903158
|
https://sentence.yourdictionary.com/equated
| 1,716,569,854,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058721.41/warc/CC-MAIN-20240524152541-20240524182541-00215.warc.gz
| 449,190,109
| 50,832
|
# Equated Sentence Examples
equated
• Of exceptional interest are the letters from Jerusalem describing the hostility of the maritime coast and the disturbances of the IIabiru (" allies "), a name which, though often equated with that of the Hebrews, may have no ethnological or historical significance s But Egypt was unable to help the loyalists, even ancient Mitanni lost its political independence, and the supremacy of the Hittites was assured.
• It has been equated on philological grounds to the Leja.
• The term "quilted" is sometimes equated with old-fashioned, but this definitely wasn't the case with the Bradley designs.
• The Greeks equated Ubasti with their Artemis, confusing her with the leonine Tafne, sister of Shoou (Apollo).
• If the form, sometimes termed a quantic, be equated to zero the n+I coefficients are equivalent to but n, since one can be made unity by division and the equation is to be regarded as one for the determination of the ratio of the variables.
• If the forms be identical the sets of symbols are ultimately equated, and the form, provided it does not vanish, is a covariant of the form ate.
• In general any pencil of lines, connected with the line a x by descriptive or metrical properties, has for its equation a rational integral function of the four forms equated to zero.
• The linear invariant a s is such that, when equated to zero, it determines the lines ax as harmonically conjugate to the lines xx; or, in other words, it is the condition that may denote lines at right angles.
• In the first class come equations in a single unknown; here the function which is equated to zero is the Y whose values for different values of X are traced, and the solution of the equation is the determination of the points where the ordinates of the graph are zero.
• The most striking coincidence is Jebel Usdum, by some equated with confidence to Sodom.
• The latter range, the Chimen-tagh, is identical in its western parts with the Piazlik-tagh and in the east must be equated with the Tsaidam chain of Przhevalsky; and it is probably continued westwards by the range which the Russian explorers call the Moscow Range or the Achik-tagh, running north of the Achik-kol and, according to Przhevalsky, connecting on the west with the Tokuz-davan.
• After a certain discount for friction and the recoil of the gun, the net work realized by the powder-gas as the shot advances AM is represented by the area Acpm, and this is equated to the kinetic energy e of the shot, in foot-tons, (I) e d2 I + p, a in which the factor 4(k 2 /d 2)tan 2 S represents the fraction due to the rotation of the shot, of diameter d and axial radius of gyration k, and S represents the angle of the rifling; this factor may be ignored in the subsequent calculations as small, less than I %.
• The mancus was equated with thirty pence, probably from the time of its introduction.
• Another remarkable indication of the decay of the ceorl's estate is afforded by the fact that in the treaties with the Danes the twihynde ceorls are equated with the Danish leysings or freedmen.
• If the inclination of the string to the vertical does not exceed a few degrees, the vertical displacement of the particle is of the second order, so that the vertical acceleration may be neglected, and the tension of the string may be equated to the gravity mg of the particle.
• Let M be the moment of the unbalanced couple required to produce the deviation; ther by equation 57, 104, the energy exerted by this couple in tht interval dt is Macit, which, being equated to the variation of energy gives da R2W da -
• On the whole, then, the work expended in producing two units of interface is 2T1+2T2-4T'12, and this, as we have seen, may be equated to 2T 12.
• The Weight Is Sometimes Equated To The Product Of The Capillary Tension (T) And The Circumference Of The Tube (27Ra), But With Little Justification.
• Sexual badinage of a heterosexual male by another such cannot be completely equated with like badinage of a heterosexual male by another such cannot be completely equated with like badinage by him of a woman.
• The Romans equated Zeus with their own supreme god Jupiter (aka Jove ).
• These terms are frequently equated with the statistical terms " quantitative " and " qualitative " .
• Rag may be roughly equated with the Western term mode or scale.
• Rogers ' and Maslow's theories of actualisation are often mistakenly equated.
• Delilah is the " Goddess night ", night being equated with the night hag, Lilith, ultimately the Babylonian Goddess Ninlil blackened.
• We believe that regionalism should not be merely equated to the provision of local news and current affairs.
• In an instant cat vomit is equated to a destroyed life for someone dealing with anxiety.
• Moreover, the Babylonian inscriptions mention the Kashshi, an Elamite race, whose name has been equated with the classical KoQaaiot, Kiauuot, and it has been held that this affords a more appropriate explanation of Cush (perhaps rather Kash), the ancestor of (the Babylonian) Nimrod in Gen.
• The latter, storm or weather god, or, in another aspect, god of rain and therefore of fertility, is specifically West Asiatic, and may be equated with Hadad and Ramman (see below).
• This cannot be equated with wilful neglect of a child.
• I actually had no idea what being a spiritual being meant and I equated it with religion at that time.
• For a long time, this was often equated with minor imperfections in craftsmanship.
• Passion doesn't need to be equated with conflict, but the little differences between you and your spouse are often responsible for that feeling of "mystery" which keeps a relationship from becoming too familiar.
• A few years ago, "boho chic" equated to hippie-inspired dresses and skirts coupled with floaty tops.
• And are soap stars still equated with actors?
• First, keep in mind that a model's super-skinny body shouldn't be equated to health.
• After the fact, this scene was interpreted as foreshadowing, as Natalie killing the rat was equated to her beating Russell in the end.
• In virtue of the mystic identity between the cosmic phenomena and sacrifice, Rita may be also viewed as the principle of the cultus; and from that sphere it passes into conduct and acquires the meaning of morality and is equated with what is " true."
| 1,464
| 6,379
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.6875
| 3
|
CC-MAIN-2024-22
|
latest
|
en
| 0.956395
|
www.allaboutpopularissues.org
| 1,696,304,379,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233511053.67/warc/CC-MAIN-20231003024646-20231003054646-00812.warc.gz
| 683,204,740
| 16,762
|
# The Great Pyramid of Giza
The Great Pyramid of Giza - The Origin
The Great Pyramid of Giza is a phenomenal site! One of the ancient wonders of the world, the Great Pyramid still produces awe and wonder! Because of its immense size, chambered design and perfect dimensions, many scholars have developed remarkable theories about its true origin and purpose.
The most popular theory is that the Great Pyramid served as a huge tomb for an ancient pharaoh of Egypt. In fact, most pyramids throughout Egypt were constructed for that purpose. However, there is now plenty of evidence to show that the Great Pyramid was a unique structure, not intended to be a traditional tomb.
The Great Pyramid of Giza - The Dimensions
The dimensions of the Great Pyramid of Giza (also known as The Great Pyramid of Khufu reveal incredible mathematical insights! The architect possessed knowledge way ahead of his time! In fact, we now know that the Great Pyramid solves one of humanity’s most difficult mathematical challenges -- the squaring of the circle. When the radius of a circle equals the height of the pyramid, then the circumference of that circle equals the perimeter of the pyramid's base. Indeed, twice the height of the Great Pyramid, divided into the distance around its base, equals nearly perfect pi. Mankind has only recently been able to calculate pi to this level of mathematical accuracy.
The length of each side of the Great Pyramid’s base is 365.2422 Royal Egyptian cubits (“pyramid cubits”). Amazingly, our astronomical year has a mean length of 365.2422454 solar days. Also, the perimeter of the base of the Great Pyramid is 36524.22 “pyramid inches,” the length of 100 years expressed in days. Calculations of such accuracy weren’t known in mainstream mathematics until the 20th century. Many additional proportional phenomena and mathematical mysteries related to the Great Pyramid have been identified, proposed and/or studied. Clearly, this is a remarkable ancient structure!
The Great Pyramid of Giza - Purpose for God?
Just as instructions were given to Moses for assembling the Tabernacle and the Ark of the Covenant, is it possible that God delivered perfect instructions for erecting the Great Pyramid of Giza? The biblical texts reveal perfect dimensions delivered to Noah for the ark and ideal proportions delivered to David for the Jewish Temple -- Like other symbols throughout the ancient scriptures, is it possible that the Great Pyramid might serve some testimonial purpose for God?
In the 8th century BC, the prophet Isaiah was speaking of the “end times” when he declared:
“In that day there will be an altar to the Lord in the midst of the land of Egypt, and a pillar to the Lord at its border” (Isaiah 19:19).
Indeed, two Egypts are mentioned in the Bible -- Upper Egypt and Lower Egypt. When the text is focused on the one unified Egypt, the Hebrew word Mazor is used. However, when Upper and Lower Egypt are referenced, as in Isaiah’s scripture, the text uses the Hebrew word Mizraim – “two Egypts.”
Remarkably, geographic studies have been done, including hydrographs and satellite imagery, confirming that the Great Pyramid of Giza is located in the direct middle of Egypt, on the “boundary” between the two ancient Egypts. In fact, Giza is Arabic for “border.” Amazingly, the Great Pyramid is a pillar on the border between these two Egypts, while also “in their midst” (the Hebrew word is betoch, meaning “between” or “within”).
The prophecy of Isaiah continues:
“And it will be for a sign and for a witness to the Lord of hosts in the land of Egypt; for they will cry to the Lord because of the oppressors, and He will send them a Savior and a Mighty One, and He will deliver them” (Isaiah 19:20).
The Great Pyramid of Giza - The Cornerstone
Throughout the ancient scriptures, Christ Jesus is compared to the chief cornerstone of a structure. Usually, the cornerstone is at the foundation of a building; however, with a pyramid, the cornerstone is at the top -- the “capstone.”
Isaiah 28:16 refers to Christ as the “precious cornerstone.” Zechariah 4:7 refers to God’s future placement of a stone at the top of his completed structure -- “Then he will bring out the capstone to shouts of 'God bless it! God bless it!'"
David also refers to the Lord in connection with a future “capstone” --
“The stone the builders rejected has become the capstone; the LORD has done this, and it is marvelous in our eyes”(Psalm 118:22-23).
Remarkably, Jesus Christ applied this same prophecy to Himself, showing that He was the “rejected stone.” Israel rejected Him as Messiah under the leadership of the Pharisees and religious teachers, so Jesus used David’s Psalm to make His dramatic point before the crowds (Matthew 21:42-44). Later, in one of Peter’s great sermons to the Jewish people, he returns to this same prophecy when speaking about the truth of Jesus Christ:
“He is 'the stone you builders rejected, which has become the capstone.' Salvation is found in no one else, for there is no other name under heaven given to men by which we must be saved”(Acts 4:11-12).
Today, it’s very interesting to note that the Great Pyramid of Giza has no capstone. In fact, there’s no historical record of any capstone ever being seen on the Great Pyramid. Something to think about as we enter “that day…”
#### Compliments of Randall Niles.
What do you think?
We have all sinned and deserve God’s judgment. God, the Father, sent His only Son to satisfy that judgment for those who believe in Him. Jesus, the creator and eternal Son of God, who lived a sinless life, loves us so much that He died for our sins, taking the punishment that we deserve, was buried, and rose from the dead according to the Bible. If you truly believe and trust this in your heart, receiving Jesus alone as your Savior, declaring, "Jesus is Lord," you will be saved from judgment and spend eternity with God in heaven.
| 1,296
| 5,924
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.984375
| 3
|
CC-MAIN-2023-40
|
latest
|
en
| 0.931655
|
https://courses.lumenlearning.com/boundless-chemistry/chapter/reaction-rates/
| 1,643,316,445,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00214.warc.gz
| 225,159,007
| 14,884
|
Measuring Reaction Rates
Reaction rates are determined by observing the changes in the concentrations of reactants or products over a specific time frame.
Learning Objectives
Produce rate expressions when given chemical reactions and discuss methods for measuring those rates
Key Takeaways
Key Points
• Reaction rate is calculated using the formula rate = Δ[C]/Δt, where Δ[C] is the change in product concentration during time period Δt.
• The rate of reaction can be observed by watching the disappearance of a reactant or the appearance of a product over time.
• If a reaction produces a gas such as oxygen or carbon dioxide, there are two ways to measure the reaction rate: using a gas syringe to measure the gas produced, or calculating the reduction in the mass of the reaction solution.
• If the reaction produces a precipitate, the amount formed can be used to determine reaction rate by measuring how long it takes for the forming precipitate to obscure the visibility of a cross through a conical flask.
Key Terms
• reaction rate: How fast or slowly a reaction takes place.
• gas syringe: An item of laboratory equipment used to withdraw a volume of gas from a closed chemical system for measurement and/or analysis.
• product: A chemical substance formed as a result of a chemical reaction.
Reaction Rate
The rate of a reaction is usually observed by watching the disappearance of a reactant or the appearance of a product within a given time period. Take the chemical reaction:
$\text{A} + 2\text{B} \rightarrow 3\text{C}$
Here, the rate of appearance of product C in time interval Δt is:
$\text{average rate}=\frac{\Delta \text{C}}{\Delta \text{t}}$
The concentration of C, [C], is usually expressed in moles/liter. This is the average rate of appearance of C during the time interval Δt. The limit of this average rate as the time interval becomes smaller is called the rate of appearance of C at time t, and it is the slope of the curve of [C] versus t at time t. This instantaneous slope, or rate, is written $\frac{\text{d}[\text{C}]}{\text{dt}}$. Since one molecule of A and two molecules of B are consumed for every three molecules of C that are produced, the rates of disappearance and appearance of these chemical species are different, but related.
Rates of disappearance and appearance of chemical species: This expression relates the rates of disappearance and appearance of chemical species in the reaction A + 2B –> 3C.
Measuring Reaction Rate
How the rate of a reaction is measured will depend on what the reaction is and what product forms. The following examples describe various ways to measure the rate of a reaction.
Reactions That Produce Gases Such as Oxygen or Carbon Dioxide
Hydrogen peroxide decomposes to produce oxygen:
$2\text{H}_2\text{O}_2(\text{aq})\rightarrow 2\text{H}_2\text{O}(\text{l})+\text{O}_2(\text{g})$
The volume of oxygen produced can be measured using the gas syringe method. The gas collects in the syringe, pushing out against the plunger. The volume of gas that has been produced can be read from the markings on the syringe. This change in volume can be converted to a change in concentration ($\Delta [\text{C}]$), and dividing this by the time of the reaction ($\Delta \text{t}$) will yield an average reaction rate.
Gas syringe method: In a reaction that produces a gas, the volume of the gas produced can be measured using the gas syringe method.
Changes in Mass
The rate of a reaction that produces a gas can also be measured by calculating the mass loss as the gas forms and escapes from the reaction flask. This method can be used for reactions that produce carbon dioxide or oxygen, but are not very accurate for reactions that give off hydrogen because the mass is too low to be accurately measured. Measuring changes in mass may also be suitable for other types of reactions.
Precipitation Reactions
In a reaction in which a precipitate is formed, the amount of precipitate formed in a period of time can be used as a measure of the reaction rate. For example, when sodium thiosulphate reacts with an acid, a yellow precipitate of sulfur is formed. This reaction is written as follows:
$\text{Na}_2\text{S}_2\text{O}_3(\text{aq})+2\text{HCl}(\text{aq})\rightarrow 2\text{NaCl}(\text{aq})+\text{SO}_2(\text{aq})+\text{H}_2\text{O}(\text{l})+\text{S}(\text{s})$
One way to estimate the rate of this reaction is to carry out the investigation in a conical flask and place a piece of paper with a black cross underneath the bottom of the flask. At the beginning of the reaction, the cross will be clearly visible when you look into the flask. However, as the reaction progresses and more precipitate is formed, the cross will gradually become less clear and will eventually disappear altogether. By using a stopwatch to time how long it takes for the cross to disappear, and then massing the amount of precipitate formed during this time, an average reaction rate can be calculated. Note that it is not possible to collect the SO2 gas that is produced in the reaction because it is highly soluble in water.
Reaction Stoichiometry
Reaction stoichiometry studies the quantitative relationships between reactants and products within a given chemical reaction.
Learning Objectives
Use stoichiometry to balance chemical equations
Key Takeaways
Key Points
• Stoichiometry comes from the Greek “stoiechion” ( element ) and “metron” (to measure). As such, stoichiometry deals with determining the amounts of reactants and products that are consumed and produced within a given chemical reaction.
• The stoichiometric coefficient of any species that does not participate in a given chemical reaction is zero.
• The principles of stoichiometry are based upon the law of conservation of mass. Matter can neither be created nor destroyed, so the mass of every element present in the product(s) of a chemical reaction must be equal to the mass of each and every element present in the reactant(s).
Key Terms
• reaction stoichiometry: Describes the quantitative relationship between reactants and products within a given chemical reaction.
• stoichiometric number: Equal to the stoichiometric coefficient in balanced equation, but positive for products (because they are produced) and negative for reactants (since they are consumed).
• stoichiometric ratio: A positive integer ratio that relates the number of moles of reactants and products involved in a chemical reaction; this ratio can be determined from the coefficients of a balanced chemical equation.
• balanced equation: When the quantity of each individual element is equal on both sides of the equation.
Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products that are consumed/produced within a given chemical reaction. In order to make any stoichiometric determinations, however, we must first look to a balanced chemical equation. In a balanced chemical equation, we can easily determine the stoichiometric ratio between the number of moles of reactants and the number of moles of products, because this ratio will always be a positive integer ratio. Consider the reaction of nitrogen gas and hydrogen gas to form ammonia (NH3):
$\text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \rightarrow 2 \text{NH}_3(\text{g})$
From the balanced equation, we can see that the stoichiometric coefficient for nitrogen is 1, while for hydrogen it is 3, and for ammonia it is 2. Therefore, the stoichiometric ratio, oftentimes referred to simply as the “mole ratio” or “molar ratio,” between N2(g), H2(g), and NH3(g) is 1:3:2. In the special case where reactants are combined in their molar ratios (in this case, 1 mole of N2(g) and 3 moles of H2(g)), they will react completely with each other, and no reactant will be left over after the reaction has run to completion. However, in most real-world situations, reactants will not combine in such perfect stoichiometric amounts. In most cases, one reactant will inevitably be the first to be completely consumed in the reaction, causing the reaction to come to a halt. This reactant is known as the limiting reactant, or limiting reagent.
From this brief description, we can see that stoichiometry has many important applications. As we will see, through balancing chemical equations and determining the stoichiometric coefficients, we will be able to determine the number of moles of product(s) that can be produced in a given reaction, as well as the number of moles of reactant(s) that will be consumed. Stoichiometry can also be used to make useful determinations about limiting reactants, and to calculate the amount of excess reactant(s) left over after a given reaction has run to completion.
The Basis of Stoichiometry
The science of stoichiometry is possible because it rests upon the law of conservation of mass. Since matter can neither be created nor destroyed, nor can a chemical reaction transform one element into another element, we can be sure that the mass of each individual element present in the reactant(s) of a given reaction must necessarily be accounted for in the product(s). This physical law is what makes all stoichiometric calculations possible. However, we can only perform these calculations correctly if we have a balanced chemical equation with which to work.
Interactive: Stoichiometry and Balancing Equations: To make hydrogen chloride or any other chemical there is only one ratio of reactants that works so that all of the hydrogen and chlorine are used to make hydrogen chloride. Try several different ratios to see which ones form a complete reaction with nothing left over. What is the simplest ratio of hydrogen to chlorine for forming hydrogen chloride?
Balancing Equations
Before performing any stoichiometric calculation, we must first have a balanced chemical equation. Take, for example, the reaction of hydrogen and oxygen gas to form liquid water:
$\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})$
As it is written here, we should notice that our equation is not balanced, because we have two oxygen atoms on the left side of the equation, but only one on the right. In order to balance this, we need to add a stoichiometric coefficient of 2 in front of liquid water:
$\text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{H}_2\text{O}(\text{l})$
In doing this, however, our hydrogens have become unbalanced. To finish balancing the equation, we must add a coefficient of 2 in front of hydrogen gas:
$2 \text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{H}_2\text{O}(\text{l})$
As we can see, the stoichiometric coefficient for any given reactant/product is the number of molecules that will participate in the reaction as written in the balanced equation. Keep in mind, however, that in our calculations, we will often be working in moles, rather than in molecules. In our example here, we can see that the stoichiometric coefficient of H2(g) is 2, while for O2(g) it is 1, and for H2O(l) it is 2. Occasionally, you might come across the term stoichiometric number, which is related to the stoichiometric coefficient, but is not the same.
Electrolysis of water: Although this image illustrates the reverse reaction of $2 \text{H}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{H}_2\text{O}(\text{l})$, the stoichiometric coefficients for each type of molecule are still the same. Water is 2, hydrogen gas is 2, and oxygen gas is 1.
For reactants, the stoichiometric number is the negative of the stoichiometric coefficient, while for products, the stoichiometric number is simply equal to the stoichiometric coefficient, remaining positive. Therefore, for our example here, the stoichiometric number for H2(g) is -2, and for O2(g) it is -1. For H2O(l), however, it is +2. This is because in this reaction, H2(g) and O2(g) are reactants that are consumed, whereas water is a product that is produced.
Lastly, you might occasionally come across some chemical species that are present during a reaction, but that are neither consumed nor produced in the reaction. A catalyst is the most familiar example of this. For such species, their stoichiometric coefficients are always zero.
Example
In the equation H2(g) + Cl2(g) → 2 HCl(g), what is the molar ratio (stoichiometric ratio) between H2(g) and HCl(g)?
In our balanced chemical equation, the coefficient for H2(g) is 1, and the coefficient for HCl(g) is 2. The molar ratio between these two compounds is therefore 1:2. This tells us that for every 1 mole of H2(g) that is consumed in the reaction, 2 moles of HCl(g) are produced.
| 2,913
| 12,651
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125
| 4
|
CC-MAIN-2022-05
|
latest
|
en
| 0.916094
|
http://perplexus.info/show.php?pid=172&cid=15073
| 1,539,915,827,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-43/segments/1539583512268.20/warc/CC-MAIN-20181019020142-20181019041642-00305.warc.gz
| 293,722,455
| 4,387
|
All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
perplexus dot info
Odd Sum (Posted on 2002-08-08)
Prove that the sum of consecutive odd numbers beginning at 1 (eg 1, 3, 5, ..) always adds up to a perfect square
Comments: ( Back to comment list | You must be logged in to post comments.)
geometric (not rigorous) solution/proof | Comment 12 of 17 |
all squares of integer size have perfect sq areas. A=s^2
starting from 1x1 square,
+
you can add 1 block horiz and 1 block vert to form this shape:
++
+
then you can add 1 corner block to complete a larger square sized 2x2
++
++
For square of size N, you can add 2N blocks(one horiz, one vertical) and 1 corner block to complete a larger square of size (N+1)
+++ ... +
+++ ... +
.
.
+++ ... +
thus starting from N=1, Square of 1x1, area=1, you can add odd number 2N+1 of blocks to form another square, where its area is a perfect square. 2N+1 is conseq sequence of numbers 3,5,7 as N increases by 1 every repetiton.
As stated by others N^2+2N+1 = (N+1)^2, but it's easier to visualize the squares.
Posted by karrio on 2004-06-08 12:06:39
Search: Search body:
Forums (0)
| 347
| 1,169
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625
| 4
|
CC-MAIN-2018-43
|
latest
|
en
| 0.820816
|
https://mathematica.stackexchange.com/questions/tagged/style?tab=active&page=2
| 1,575,991,740,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00328.warc.gz
| 457,397,412
| 39,715
|
# Questions tagged [style]
Questions about applying a consistent look or form to Mathematica output, including use of the built-in styling constructs (Style, PlotStyle, etc). For questions specific to styling in notebooks with stylesheet properties please use the [stylesheet] tag.
274 questions
Filter by
Sorted by
Tagged with
915 views
### Multiple GridLines of different colors
I wanna plot 2 gridlines in colors. One in red. Other in blue. How to do it? Wrong code: ...
80 views
### Adding overlapping non-weighted, directed edges to a weighted, undirected graph
I have the following Mathematica code based on the GDB1 instance from the CARP literature: ...
184 views
### Weighted graph with multiple different coloured non-weighted paths - styling
This follows on from my previous question answered here. Ultimately, I now have a graph which I've simplified for posting here: ...
46 views
### Axes position and axes label in RevolutionPlot [duplicate]
Consider Revolution plot: ...
678 views
### Changing error bars color
I have the following plot ...
80 views
### How to automate creation of sections in mathematica worksheets?
I want to do a for loop to compute some values, create a new (sub-)section and print those values there within each iteration. How would it be possible?
30 views
### Labels on 2D ListPlot Markers [duplicate]
If we take the example ListPlot[{{1.5, 2}, {1, -2}, {2.5, 6}, {3, -5}} -> {"a", "b", "c", "d"}] I would to see the labels correctly correspond to the 2D ...
133 views
### Change distance between vertical y-axis label and plot
I'm trying to do something that should be fairly straightforward: having a vertical y-axis label. My problem is, it creates a ridiculously large space between the label and the axis (exactly what I'm ...
58 views
### How to add boundaries for the points of a ListPlot?
I have a plot as below: ListPlot[{{{1, 5}}, {{4, 2}}}, PlotStyle -> {{Red, PointSize[0.03]}, {Green, PointSize[0.03]}}, PlotRange -> {{0, 5}, {0, 6}}] I ...
83 views
### How can I use the help document style？
I have spent a lot of time looking for this information on the Internet。Unfortunately,I can not find it。Please help me。Thank you，Thank you。 （I do not how to describe this question。May god forgive me ...
257 views
### Syntax highlighting in text cells
I want to write some documentation for a notebook using text cells. Now, I can select parts of the text within such a text cell and set its style to "Input". This does change its font, but there is no ...
111 views
### Graph (network) styling: avoid vertices and edges overlapping
Is there a way to avoid the edges and vertices 'overlapping' in a graph with custom styling? That is, the edges should stop once they reach the 'border' of the shape representing the vertex (in this ...
113 views
### Preserving changed Output style
Let's say I have a stylesheet with two slightly different "Output" and "Output2" styles. ...
74 views
### How to make the legend with color more compact in a plot?
I have a plot as below:- ...
70 views
### Highlight selected text everywhere?
Notebook++ has a very useful feature which highlights all occurrences of certain text once it has been selected with the mouse cursor in one spot: It is very useful in order to jump between different ...
21 views
### Match axes lengths of two plots? [duplicate]
Consider the two plots ...
41 views
### Adding Legends in the plot, aligned in a row
How can I add the legend in a plot, INSIDE the plot (as in the figure below), but in a way that they are arranged in row and not in column? What I have done so far: ...
481 views
### Replicating a PlotStyle from Trees, Maps, and Theorems
How might I make a plot with this style, where the callouts are on the side and the y-axes chopped off for a minimal look? Here's some example data with the callout that I'd like to add, but I'm not ...
50 views
### Make size of ListPolarPlot with PolarAxes consistent
I am currently having an issue with ListPolarPlot, specifically with achieving a consistent size among multiple plots. Consider the following illustration of my ...
85 views
### Graphics3D ignores EdgeForm
Works great: Graphics[{FaceForm[White], EdgeForm[AbsoluteThickness[10]], Polygon[{{0, 0}, {1, 0}, {0, 1}}]}] Ignores color and thickness: ...
57 views
### Docked Cell Slider that controls magnification of Output Cells only
The following code (found here on stack exchange) creates a docked cell which contains a slider that can be used to change the magnification of the entire notebook. ...
147 views
### How can I set the style of one Output cell so it doesn't change after re-evaluating its Input cell?
I have created a notebook for my classes which is full of inputs that need to be evaluated by my students. All these inputs produce standard outputs, which are fine as they are. But there is one of ...
73 views
### Customize plot with CapForm?
Simple plotting question: I'd like to keep the base style of the "Business" theme, but with two slight modifications: Use CapForm to round the ends of my plot ...
335 views
### How to increase dash length in DotDashed
How can I modify the usual curve style DotDashed such that the length of the dashed portion is increased or is controllable? For example in ...
179 views
### How do I reset Mathematica's DisplayFormulaNumbered cell style numbering?
Since I am writing many equations, in various chapters, I need a way to reset numbering in this cell style. The following shows what I need: x + y = 15 (1) x - z = ...
57 views
### Cell style inheritance and specifying the same style twice [duplicate]
Bug introduced in V11.1 or earlier and persists through V11.3 [CASE:4061571], confirmed Consider the following three examples of a cell: ...
122 views
### Changing line style in Dendrogram
How does one change the line style in Dendrogram? BaseStyle changes the style on the points (vertexes) and ...
504 views
### How to create the waved style for curves in Graphics?
There are the dashed and dotted styles for lines and circles when we plot them with Graphics....
272 views
### How to specify custom graph styles for GraphStyle?
There is a way to specify custom themes for plots (Is it possible to define a new PlotTheme?), but themes do not apply to Graph objects, not even to ...
209 views
### Color in TableForm
Happy Easter! How can I change the background color of one specific row in TableForm? Or can I even do it ? for instance: ...
27 views
### How to print expression in color using Print function? [duplicate]
I am wondering how would you print the expression with color using function Print. For example, I tried the code below but the output of print function is just ...
38 views
Writing: ...
56 views
### Merging list valued style options
As shown in Difference between ParentList and Inherited we can use ParentList to nicely merge additional rules. It seems to work nicely when it comes to inheriting ...
229 views
### Multiple colors on a single graph edge?
Is there any way to put two colors on one single graph edge? For example, I want to highlight two cyclic paths (red and blue). But, if I put the two colorings into one image, it does not work as I ...
93 views
### Why does Mathematica ignore my stylesheet for DisplayFormulaNumbered cells?
I made a custom stylesheet, installed it, and applied it to a new blank notebook. Adding a cell and changing it to Text results in the expected behavior. The font and font size change to whatever the ...
47 views
### How to style x-label of two data-sets differently in FrameLabel?
Hi I have following code, unfortunately it does not lead to the result of the label G1 shown in black and the label G3 shown in red with fontsize 22. ...
534 views
### How to rescale x axis
Please suggest how to rescale x axis by t0: I am trying to plot given below exponential. ...
75 views
...
275 views
### How to reverse the x axis in ContourPlot
I have the following equation p = 1.6; α = 0.001; r = 0.6; η = 0.04; ω = 1; R ω p Sin[ω τ] + R ω p α - 9/4 r p R^3 ω - η p R == 0 I want to plot ...
2k views
### FillingStyle and Opacity won't work together in Mathematica
I try to plot some data in Mathematica using the ListLinePlot[]. Unfortunately the plot option ColorFunction, ...
3k views
### How can I create a help button that looks more “professional”?
My goal, is to make the following button, ...
749 views
### wolframscript — How to print in color
I want to print to the console in color from a wolframscript. I tried just using a Style: ...
125 views
### How to highlight the cells that satisfies the condition?
MemberQ[IntegerDigits [#], 3] & /@ Range[100] Grid[ArrayReshape[Range[100], {10, 10}], Frame -> All] I want to get a 10 by 10 grid, that highlights all ...
422 views
### Is there a way to “hold” prefix / infix / postfix notation?
Is it possible to preserve a specific style of notation in $Mathematica$ output? foo // bar (* bar[foo] *) Hold[foo // bar] (* bar[foo] *) I'm writing some code ...
143 views
### Better aesthetics for graph vertices?
This is a follow-up question to: Unable to export a formatted graph? I trying to layout a graph so that if you zoom in on a vertex, things look nice. To see what I mean by nice, let me just describe ...
154 views
### How to make a discrete ListPlot3D look like a Histogram3D?
I want to plot a list of 3D points and make it look like this: where each value is represented by the height of a rectangular parallelepiped (with square top and bottom face). As a simple example I'...
58 views
### Clarifying which segement of a curve overlap is in the foreground/background [duplicate]
I plotted the left curve using regular ParametricPlot3D. In parts where the curve overlaps itself, it is unclear which segment is in the foreground and which in the ...
238 views
### Setting each “curve” on a 3D plot to be of different colour similar to that of its 2D plot
Consider the functions ...
| 2,363
| 9,979
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.609375
| 3
|
CC-MAIN-2019-51
|
latest
|
en
| 0.883264
|
https://www.earlymathca.org/copy-of-activities-1
| 1,680,085,808,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948965.80/warc/CC-MAIN-20230329085436-20230329115436-00582.warc.gz
| 819,855,834
| 148,588
|
top of page
## Dominoes
Dominoes are a great resource for learning numbers and patterns. They can be used for matching, counting, adding, subtracting, subitizing (recognizing how many are there without counting), comparing, and more!
Children can learn counting with dominoes by placing counters over each dot as they count. Some things to use as counters:
• Buttons
• Small pom pons
• Shapes cut from colored paper
• Beans
• Small rocks
• Counters
Dominoes can also be used for learning the skill known as “counting on." Children use this skill when learning addition. For example, a student wants to add 5 plus 3. They do not yet know the answer so they figure it out by counting. “Counting all” involves starting by counting out five, then counting out three and then counting all objects. “Counting on” involves starting with five and then counting three more without starting over. The child would say, "fiiiive... six, seven, eight, there are eight all together."
Domino Activities and Template
bottom of page
| 238
| 1,030
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.828125
| 3
|
CC-MAIN-2023-14
|
latest
|
en
| 0.950411
|
https://drexel28.wordpress.com/2010/12/27/interesting-combinatorial-sum-involving-inversions/
| 1,529,756,593,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267864958.91/warc/CC-MAIN-20180623113131-20180623133131-00085.warc.gz
| 584,879,353
| 20,945
|
# Abstract Nonsense
## Interesting Combinatorial Sum Involving Inversions
Point of post: In this post we’ll solve the compute sum
$\displaystyle \sum_{\pi\in S_n}q^{\text{inv}(\pi)}$
where $q\in\mathbb{C}$ and $\text{inv}(\pi)$ is the number of inversions of a permutation. In doing so we find an interesting way to prove that $\#\left(S_n\right)=n!$.
Theorem: Let $n\in\mathbb{N},\text{ }n\geqslant 2$, then
$\displaystyle \sum_{\pi\in S_n}q^{\text{inv}(\pi)}=\prod_{k=1}^{n}\sum_{r=0}^{k-1}q^r$
Proof: We proceed by induction. For $n=2$ we see that $S_2=\{\text{id}_{[2]},(1,2)\}$ and thus
\displaystyle \begin{aligned}\sum_{\pi\in S_2}q^{\text{inv}(\pi)} &= q^{\text{inv}(\text{id}_{[2]})}+q^{\text{inv}((1,2))}\\ &= q^0+q^1\\ &= 1+q\end{aligned}
as claimed. Next, suppose that this is true for $n$. Consider then $S_{n+1}$ is partitioned into two blocks: those which fix $n+1$ and those that don’t, call these blocks $B_1$ and $B_2$ respectively. Identify each element of $S_{n+1}$ with its associated $n+1$-tuple. Notice then that the elements of $B_1$ are of the form $(\ast,\cdots,\ast,n+1)$ and the elements of $B_2$ are of the form $(\ast,\cdots,\ast,n+1,\cdots,\ast)$. For each element of $B_1$ the associated $n+1$-tuple can be thought of as the association of an $n$-tuple with an element of $S_n$ with the addition of the last coordinate, always being $n+1$. In particular, the elements of $B_1$ are precisely the permutation of the form $\sigma:[n+1]\to[n+1]$ such that $\sigma_{\mid [n]}\in S_n$ and in particular $\text{inv}(\sigma)=\text{inv}\left(\sigma_{\mid [n]}\right)$. It evidently follows then that
$\displaystyle \sum_{\pi\in B_1}q^{\text{inv}(\pi)}=\sum_{\pi\in S_n}q^{\text{inv}(\pi)}$
Next note that if $\sigma\in B_2$ and the associated $n+1$-tuple is
$(\underbrace{\ast,\cdots,\ast}_{\ell},n+1,\cdots,\ast)\quad\mathbf{(1)}$
Then, if one removes the coordinate which $n+1$ resides in then the resulting tuple can be uniquely identified with an element of $S_n$. Moreover, if $\pi$ is that associated permutation then $\text{inv}(\sigma)=\text{inv}(\pi)+\ell$. It follows then that for each fixed $\ell$ we call $B_{2,\ell}$ all the elements of $B_2$ of the form $\mathbf{(1)}$ then
$\displaystyle \sum_{\pi\in B_{2,\ell}}q^{\ell}\sum_{\pi\in S_n}$
and thus noticing that
$B_2=B_{2,1}\sqcup\cdots\sqcup B_{2,n}$
it follows that
$\displaystyle \sum_{\pi\in B_2}q^{\text{inv}(\pi)}=\sum_{\ell=1}^{n}q^{\ell}\sum_{\pi\in S_n}q^{\text{inv}(\pi)}$
So, we may conclude that
\displaystyle \begin{aligned}\sum_{\pi\in S_{n+1}}q^{\text{inv}(\pi)} &= \sum_{\pi\in B_1}q^{\text{inv}(\pi)}+\sum_{\pi\in B_2}q^{\text{inv}(\pi)}\\ &= \sum_{\pi\in S_n}q^{\text{inv}(\pi)}+\sum_{\ell=1}^{n}q^{\ell}\sum_{\pi\in S_n}q^{\text{inv}(\pi)}\\ &= \left(\sum_{\pi\in S_n}q^{\text{inv}(\pi)}\right)\left(\sum_{\ell=0}^{n}q^\ell\right)\\ &= \left(\prod_{k=1}^{n}\sum_{r=0}^{n-1}q^r\right)\left(\sum_{r=0}^{n}q^r\right)\\ &= \prod_{k=1}^{n+1}\sum_{r=0}^{k-1}q^r\end{aligned}
and the induction is complete. $\blacksquare$
From this we get an interesting way to prove that $\#\left(S_n\right)=n!$. Indeed, letting $q=1$ in the above gives
$\displaystyle \sum_{\pi\in S_n}1=\prod_{k=1}^{n}k=n!$
| 1,241
| 3,217
|
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 49, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375
| 4
|
CC-MAIN-2018-26
|
latest
|
en
| 0.773849
|
https://www.theguardian.com/science/2020/sep/07/can-you-solve-it-the-smallest-biggest-triangle-in-the-world
| 1,600,662,461,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2020-40/segments/1600400198887.3/warc/CC-MAIN-20200921014923-20200921044923-00140.warc.gz
| 1,131,596,319
| 41,283
|
# Can you solve it? The smallest biggest triangle in the world
The Earth ain’t big enough for this tiny shape
Today’s puzzle is simple and spectacular. It asks you to construct a triangle whose existence seems to defy reason.
Show that there is a triangle, the sum of whose three heights is less than 1mm, that has an area greater than the surface of the Earth (510m km2).
Instinctively, one feels that such a triangle is impossible. How can something that is so arbitrarily tiny be so arbitrarily big? Yet it is possible to describe such a triangle using the tools of elementary geometry.
Don’t be put off by the terminology here. There are no hidden catches.
The area of a triangle is half multiplied by the base multiplied by the height.
The height of a triangle is the perpendicular distance from a side (or an extension of that side) to the opposing vertex. Since a triangle has three sides, it has three heights.
The puzzle was devised by Trần Phương, a legendary maths educator from Vietnam. It was originally set in the final round of a Vietnamese TV gameshow for smart teenagers in 1998. (None of them solved it.)
Trần has written more than 50 maths books and for the last three decades has trained Vietnam’s Mathematical Olympiad teams. (On average, Vietnam outperforms the UK.) Trần is also a songwriter. He wrote the Covid-19 song below earlier in the year to encourage people to stay safe.
Although do share ways you might want to tackle the problem, and also feel free discuss your favourite triangles.
I’ll be back at 5pm UK time with the solution and a discussion.
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
I’m the author of several book of maths and puzzles, and also co-author of the children’s book series Football School. The latest book in the series, Epic Heroes, which tells the 50 best stories in football, is out on October 1 and available for preorder here.
| 430
| 1,993
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125
| 4
|
CC-MAIN-2020-40
|
latest
|
en
| 0.965058
|
https://www.tensorflow.org/versions/r1.15/api_docs/python/tf/random/uniform_candidate_sampler?hl=bg
| 1,643,396,046,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320306335.77/warc/CC-MAIN-20220128182552-20220128212552-00696.warc.gz
| 1,071,377,051
| 58,144
|
Help protect the Great Barrier Reef with TensorFlow on Kaggle
# tf.random.uniform_candidate_sampler
Samples a set of classes using a uniform base distribution.
This operation randomly samples a tensor of sampled classes (sampled_candidates) from the range of integers [0, range_max).
The elements of sampled_candidates are drawn without replacement (if unique=True) or with replacement (if unique=False) from the base distribution.
The base distribution for this operation is the uniform distribution over the range of integers [0, range_max).
In addition, this operation returns tensors true_expected_count and sampled_expected_count representing the number of times each of the target classes (true_classes) and the sampled classes (sampled_candidates) is expected to occur in an average tensor of sampled classes. These values correspond to Q(y|x) defined in this document. If unique=True, then these are post-rejection probabilities and we compute them approximately.
true_classes A Tensor of type int64 and shape [batch_size, num_true]. The target classes.
num_true An int. The number of target classes per training example.
num_sampled An int. The number of classes to randomly sample. The sampled_candidates return value will have shape [num_sampled]. If unique=True, num_sampled must be less than or equal to range_max.
unique A bool. Determines whether all sampled classes in a batch are unique.
range_max An int. The number of possible classes.
seed An int. An operation-specific seed. Default is 0.
name A name for the operation (optional).
sampled_candidates A tensor of type int64 and shape [num_sampled]. The sampled classes, either with possible duplicates (unique=False) or all unique (unique=True). In either case, sampled_candidates is independent of the true classes.
true_expected_count A tensor of type float. Same shape as true_classes. The expected counts under the sampling distribution of each of true_classes.
sampled_expected_count A tensor of type float. Same shape as sampled_candidates. The expected counts under the sampling distribution of each of sampled_candidates.
[{ "type": "thumb-down", "id": "missingTheInformationINeed", "label":"Missing the information I need" },{ "type": "thumb-down", "id": "tooComplicatedTooManySteps", "label":"Too complicated / too many steps" },{ "type": "thumb-down", "id": "outOfDate", "label":"Out of date" },{ "type": "thumb-down", "id": "samplesCodeIssue", "label":"Samples / code issue" },{ "type": "thumb-down", "id": "otherDown", "label":"Other" }]
[{ "type": "thumb-up", "id": "easyToUnderstand", "label":"Easy to understand" },{ "type": "thumb-up", "id": "solvedMyProblem", "label":"Solved my problem" },{ "type": "thumb-up", "id": "otherUp", "label":"Other" }]
| 610
| 2,744
|
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 2.515625
| 3
|
CC-MAIN-2022-05
|
latest
|
en
| 0.73108
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.